🧠:Okay, so I need to figure out the value of the sum (sum_{k-1}^{49}(-1)^{k}binom{99}{2k}). Wait, hold on, the summation limits are a bit confusing. It says (k-1) to 49. Hmm, maybe that's a typo? Probably meant (k=1) to 49. Yeah, that makes more sense. So the summation is from (k=1) to (k=49) of ((-1)^k times binom{99}{2k}). The answer choices are in terms of powers of 2, so there's likely a combinatorial identity or generating function approach here.First, let me recall some binomial theorem and combinatorial identities. The standard binomial theorem tells us that ((1 + x)^n = sum_{k=0}^{n} binom{n}{k}x^k). But here, we have a sum that involves only even indices (2k) and alternating signs. So maybe considering the expansion of ((1 + x)^{99}) and then plugging in a specific value of x to pick out the even terms?Wait, also, there's an alternating sign ((-1)^k). So perhaps if I set x to some value such that when combined with the even terms, the alternating sign comes into play. Let me think.I know that the sum of the even binomial coefficients can be found using ((1 + 1)^{n} + (1 - 1)^{n}) divided by 2, which gives (2^{n-1}). Similarly, the sum of the odd ones is (2^{n-1}). But here we have an alternating sum, not just the sum. So maybe using complex numbers?Alternatively, consider the generating function for the even terms. For even indices, we can use ((1 + x)^{99} + (1 - x)^{99}) over 2. Then substituting x with something. But here, each term is multiplied by ((-1)^k). So if I let x be a value that when squared gives -1? Hmm, maybe x = i, where i is the imaginary unit.Let me write down the generating function:[sum_{k=0}^{49} binom{99}{2k} (-1)^k]Wait, the original sum is from k=1 to 49, but if I start from k=0, then subtract the k=0 term. Let me check: when k=0, the term is ((-1)^0 times binom{99}{0}) = 1. So the original sum S = (sum from k=0 to 49) - 1. So maybe first compute the sum from k=0 to 49 and then subtract 1.But let's see. Let's define S = (sum_{k=0}^{49} (-1)^k binom{99}{2k}). Then the original problem's sum is S - 1. So let's compute S first.To compute S, note that S is the sum over even indices of (binom{99}{2k} (-1)^k). Let's relate this to the generating function.Recall that (sum_{k=0}^{n} binom{n}{k} x^k = (1 + x)^n). If we want only even k, we can use (frac{(1 + x)^n + (1 - x)^n}{2}). So, substituting x with something. But here, each term is multiplied by ((-1)^k). So, maybe if we set x such that x^{2k} becomes (-1)^k. That would mean x^2 = -1, so x = i or x = -i. Let me check.Let’s try substituting x = i into the generating function. Then:[(1 + i)^{99} + (1 - i)^{99} = 2 sum_{k=0}^{49} binom{99}{2k} i^{2k} = 2 sum_{k=0}^{49} binom{99}{2k} (-1)^k]So this sum S is equal to (frac{(1 + i)^{99} + (1 - i)^{99}}{2}). Therefore, S = (frac{(1 + i)^{99} + (1 - i)^{99}}{2}). Then the original problem's sum is S - 1.So I need to compute ((1 + i)^{99} + (1 - i)^{99}), divide by 2, subtract 1, and then see what that is.First, let's compute (1 + i) and (1 - i) in polar form. The modulus of (1 + i) is (sqrt{2}), and the argument is (pi/4). Similarly, (1 - i) has modulus (sqrt{2}) and argument (-pi/4).Therefore, ((1 + i)^{99} = (sqrt{2})^{99} times e^{i times 99 times pi/4}).Similarly, ((1 - i)^{99} = (sqrt{2})^{99} times e^{-i times 99 times pi/4}).Adding these two together:[(sqrt{2})^{99} left( e^{i times 99pi/4} + e^{-i times 99pi/4} right) = 2 (sqrt{2})^{99} cos(99pi/4)]Therefore, S = (frac{2 (sqrt{2})^{99} cos(99pi/4)}{2}) = ((sqrt{2})^{99} cos(99pi/4)).Simplify the cosine term. Let's compute the angle 99π/4. Let's reduce the angle modulo 2π. 99 divided by 8 is 12.375, so 12 times 2π is 24π, which is 12 full circles. 99π/4 = 24π + 3π/4. Therefore, 99π/4 is equivalent to 3π/4 in terms of the unit circle. Therefore, cos(99π/4) = cos(3π/4) = -√2/2.So S = ((sqrt{2})^{99} times (-sqrt{2}/2)).Simplify this:First, ((sqrt{2})^{99} = (2)^{99/2}). Multiply by (-sqrt{2}/2):[(2)^{99/2} times (-sqrt{2}/2) = - (2)^{99/2} times 2^{1/2} / 2 = - 2^{(99/2 + 1/2)} / 2 = -2^{50}/2 = -2^{49}]Therefore, S = (-2^{49}). Then the original sum is S - 1 = (-2^{49} - 1). Wait, but the answer choices are -2^50, -2^49, 2^49, 2^50. Wait, but hold on, the original problem says "the value of (sum_{k-1}^{49}(-1)^{k}binom{99}{2k})". If that is indeed from k=1 to 49, then the sum is S - 1, where S = -2^{49}, so the total would be -2^{49} - 1. But that's not among the options. So maybe I made a mistake here.Wait, let me double-check the steps. First, when I considered S as the sum from k=0 to 49 of (-1)^k binom{99}{2k}, which is equal to [(1 + i)^99 + (1 - i)^99]/2. Then computed that as -2^{49}. Then the original problem's sum is S - 1, which would be -2^{49} - 1, but that's not an answer choice. So something's wrong here.Wait, but hold on, the upper limit is 49. But when n=99, the maximum possible k for 2k is 49.5, but since k is integer, the maximum is 49. Therefore, the sum from k=0 to 49 of (-1)^k binom{99}{2k} is S, which we found as -2^{49}. But the original sum is from k=1 to 49, so subtract the term at k=0, which is (-1)^0 binom{99}{0} = 1. So the original sum is S - 1 = -2^{49} - 1. But the answer options don't include -2^{49} -1. So maybe my initial assumption was wrong.Alternatively, perhaps the problem is written incorrectly. Maybe the summation is from k=0 to 49? If that were the case, then the answer would be -2^{49}, which is option B. But the problem states from k-1 to 49, which is likely a typo. Alternatively, if it's from k=1 to 49, then maybe the answer is -2^{49} -1, but that's not listed. So perhaps my mistake is in the calculation of S.Wait, let's re-examine the step where I calculated cos(99π/4). 99 divided by 4 is 24.75, which is 24 + 3/4. So 24π + 3π/4. Since cosine has period 2π, subtracting 12 full circles (24π), so cos(99π/4) = cos(3π/4) = -√2/2. That part seems correct.Then S = (sqrt(2))^{99} * (-sqrt(2)/2) = (2)^{99/2} * (-2^{1/2}/2) = -2^{(99 +1)/2}/2 = -2^{100/2}/2 = -2^{50}/2 = -2^{49}. That's correct.Therefore, S = -2^{49}, so if the problem's sum is from k=0 to 49, the answer is -2^{49}, which is option B. But the problem says from k-1 to 49. If it's from k=1 to 49, then the answer is S - 1 = -2^{49} - 1, which is not among the options. Therefore, there must be a mistake either in the problem statement or in my approach.Alternatively, maybe the summation is over k=1 to 49, but the upper limit of 2k is 98, which is even, so 2k=98 implies k=49. So that's okay.Wait, another approach: Maybe using generating functions differently. Let's think about the binomial expansion of (1 - 1)^{99} and (1 + 1)^{99}. But (1 -1)^{99} is 0, and (1 +1)^{99} is 2^{99}. But how does that relate to the sum of even terms with alternating signs?Alternatively, consider the generating function f(x) = sum_{k=0}^{99} binom{99}{k} x^k = (1 + x)^{99}. Now, the sum we want is sum_{k=1}^{49} (-1)^k binom{99}{2k}. Let's denote j = 2k, so j ranges from 2 to 98 even numbers. Then the sum becomes sum_{j=2}^{98 even} (-1)^{j/2} binom{99}{j}.Alternatively, write j = 2k, so sum_{j=0}^{98} (-1)^{j/2} binom{99}{j} where j is even. Then subtract the terms for j=0 and j=98 if necessary. Wait, but (-1)^{j/2} is equivalent to (-1)^{k} when j=2k.But maybe another substitution. Let’s consider evaluating the generating function at x = i, since x^2 = -1. Let’s compute f(i) = (1 + i)^{99} = sum_{k=0}^{99} binom{99}{k} i^k. Similarly, f(-i) = (1 - i)^{99}.Then, consider [f(i) + f(-i)] / 2 = sum_{k=0}^{49} binom{99}{2k} (-1)^k. Because i^{2k} = (-1)^k. Therefore, this is exactly S, the sum from k=0 to 49 of (-1)^k binom{99}{2k} which equals [(1 + i)^99 + (1 - i)^99]/2, which we computed as -2^{49}.Therefore, if the problem's sum is from k=0 to 49, the answer is -2^{49}, which is option B. But the problem's summation is written as from k-1 to 49, which is probably a typo, and intended to be k=0 to 49. Alternatively, if the original problem actually starts at k=1, then the answer would be -2^{49} -1, which isn't an option, so likely the problem has a typo, and the correct answer is B.Alternatively, maybe I miscalculated something. Let me check once more:Compute (1 + i)^99:(1 + i) has magnitude sqrt(2), angle pi/4. So (1 + i)^99 = (sqrt(2))^99 * e^{i * 99 * pi/4}.Similarly, (1 - i)^99 = (sqrt(2))^99 * e^{-i * 99 * pi/4}.Adding them: (sqrt(2))^99 [e^{i * 99pi/4} + e^{-i *99pi/4} ] = 2*(sqrt(2))^99 * cos(99pi/4).Compute 99pi/4: 99 divided by 4 is 24.75, so subtract 12 full circles (24pi) to get 99pi/4 - 24pi = 99pi/4 - 96pi/4 = 3pi/4. So cos(3pi/4) = -sqrt(2)/2.Therefore, the sum is 2*(sqrt(2))^99*(-sqrt(2)/2) = - (sqrt(2))^{100} = - (2^{1/2})^{100} = -2^{50}. Wait, hold on! Wait, (sqrt(2))^{100} is 2^{50}, so multiplying by -1 gives -2^{50}. But earlier I had:2*(sqrt(2))^99 * (-sqrt(2)/2) = - (sqrt(2))^{99} * sqrt(2) = - (sqrt(2))^{100} = -2^{50}Wait, so previously, I thought that S = -2^{49}, but now this calculation gives S = -2^{50}?Wait, there's an inconsistency here. Let's recast:Starting with [ (1 + i)^99 + (1 - i)^99 ] / 2 = SWe have (1 + i)^99 + (1 - i)^99 = 2 * (sqrt(2))^99 * cos(99pi/4)But cos(99pi/4) = cos(3pi/4) = -sqrt(2)/2.Therefore, 2 * (sqrt(2))^99 * (-sqrt(2)/2) = - (sqrt(2))^{100} = -2^{50}Then S = [ -2^{50} ] / 2 = -2^{49}Wait, no! Wait, the entire expression (1 + i)^99 + (1 - i)^99 equals 2 * (sqrt(2))^99 * cos(99pi/4) = 2 * (sqrt(2))^99 * (-sqrt(2)/2) = - (sqrt(2))^{100} = -2^{50}Therefore, S = [ (1 + i)^99 + (1 - i)^99 ] / 2 = (-2^{50}) / 2 = -2^{49}Therefore, S is indeed -2^{49}. Therefore, if the sum is from k=0 to 49, the answer is -2^{49}, which is option B. But the original problem says from k-1 to 49. If it's k=1 to 49, then the sum is S - [term at k=0] = -2^{49} -1. But since this is not an option, and given the answer choices include -2^{49}, likely the problem intended the sum from k=0 to 49, and the typo is in the limits. Therefore, the answer is B, -2^{49}.But just to make sure, let's check with small n to see if the formula holds.For example, take n=2. Then sum_{k=0}^{1} (-1)^k binom{2}{2k} = (-1)^0 binom{2}{0} + (-1)^1 binom{2}{2} = 1 -1 = 0. According to the formula, S = -2^{(2)/2} = -2^1 = -2. But this doesn't match. Wait, this is a problem. So there's a mistake here.Wait, when n=2, sum_{k=0}^1 (-1)^k binom{2}{2k} = 1 -1 = 0. But according to the formula we derived, S = -2^{n/2} when n is even? Wait, but n=99 in the original problem. Wait, maybe the formula is different. Let's check for n=2.Compute (1 + i)^2 + (1 - i)^2. (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i -1 = 2i. (1 - i)^2 = 1 - 2i + i^2 = 1 -2i -1 = -2i. So sum is 2i + (-2i) = 0. Then S = 0 / 2 = 0. Which matches the actual sum. According to our previous formula, n=2, S = -2^{2/2} = -2^1 = -2, which is wrong. So the formula is not generalizable as S = -2^{n/2} for even n. Therefore, my mistake was assuming a general formula, but for n=99, which is odd.Wait, n=99 is odd. Let me see. Let's take n=3. Sum from k=0 to 1 (since 2k <=3, k=0 and 1). Sum is (-1)^0 binom{3}{0} + (-1)^1 binom{3}{2} = 1 -3 = -2. According to the formula:(1 + i)^3 + (1 - i)^3 divided by 2. Compute (1 + i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i -3 -i = -2 + 2i. (1 - i)^3 = 1 - 3i + 3i^2 -i^3 = 1 - 3i -3 +i = -2 -2i. Sum is (-2 + 2i) + (-2 -2i) = -4. Divided by 2 is -2. Which matches. Then according to our previous calculation:( sqrt(2) )^3 * cos(3pi/4 * 3). Wait, no, for n=3, the angle would be 3pi/4 *3?Wait, maybe the general formula is that for n odd, the sum is -2^{(n-1)/2} * sqrt(2) ?Wait, no, let's recast. For n=99:(1 + i)^n + (1 - i)^n = 2*(sqrt(2))^n * cos(n pi/4)Therefore, for n=99:cos(99pi/4) = cos(24pi + 3pi/4) = cos(3pi/4) = -sqrt(2)/2.Thus, 2*(sqrt(2))^{99}*(-sqrt(2)/2) = - (sqrt(2))^{100} = -2^{50}Therefore, [ (1 + i)^{99} + (1 - i)^{99} ] / 2 = -2^{50}/2 = -2^{49}. So S = -2^{49}But in the case of n=3:(1 + i)^3 + (1 - i)^3 = -4, so divided by 2 is -2, which is equal to -2^{ (3+1)/2 } = -2^{2} = -4? Wait no, for n=3, our formula gives S = -2^{(3)/2} * something?Wait, perhaps my confusion is arising from different n. In the original problem, n=99, so when we calculate (1 + i)^n + (1 - i)^n, it's 2*(sqrt(2))^n * cos(n pi/4). Then, divided by 2 gives S = (sqrt(2))^n * cos(n pi/4).But for n=3:( sqrt(2) )^3 * cos(3pi/4) = (2^{1.5}) * (-sqrt(2)/2) = 2^{1.5} * (-2^{-0.5}) = -2^{1.5 -0.5} = -2^1 = -2. Which matches.For n=2:( sqrt(2) )^2 * cos(2pi/4) = 2 * cos(pi/2) = 2*0=0. Which matches.For n=1:( sqrt(2) )^1 * cos(pi/4) = sqrt(2) * (sqrt(2)/2) = 1. Which matches the sum from k=0 to 0: (-1)^0 * binom(1}{0} =1.So the formula works. Therefore, for n=99:S = (sqrt(2))^{99} * cos(99pi/4) = (sqrt(2))^{99} * (-sqrt(2)/2) = - (sqrt(2))^{100}/2 = -2^{50}/2 = -2^{49}Thus, the sum from k=0 to 49 is -2^{49}. If the problem's sum is from k=1 to 49, then the answer is -2^{49} -1. But since that is not an option, and option B is -2^{49}, it's likely that there was a typo in the summation limits, and the correct answer is B.Alternatively, maybe the problem is correct as written, but I missed something. Wait, another approach: expand (1 - 1)^{99} and (1 + 1)^{99} and relate to the sum.But (1 - 1)^{99} = 0, (1 +1)^{99} = 2^{99}. Also, the sum of even binomial coefficients is 2^{98} for n=99, but with alternating signs.Alternatively, consider that the required sum is the real part of (1 + i)^{99}, but scaled. Wait, we already did that.Alternatively, use generating function for even indices with alternating signs.Let me consider generating function G(x) = sum_{k=0}^{∞} binom{99}{2k} (-1)^k x^{2k}. Then setting x=1 gives our desired sum. But G(x) can be expressed as [ (1 + sqrt(-x))^{99} + (1 - sqrt(-x))^{99} ] / 2. Setting x=1, we get [ (1 + i)^{99} + (1 - i)^{99} ] / 2, which is the same as before.Thus, confirming the result is -2^{49}. Hence, the answer is B, -2^{49}, assuming the summation starts at k=0. Since the original problem may have a typo, but given the answer choices, it's safe to go with B.Final Answerboxed{B}

✅:To determine the value of the sum (sum_{k=1}^{49}(-1)^{k}binom{99}{2k}), we start by considering the sum from (k=0) to (49) and then subtract the term for (k=0).First, we use the generating function approach. The sum of even-indexed binomial coefficients with alternating signs can be related to the complex numbers (1 + i) and (1 - i). Specifically, we use the identity:[sum_{k=0}^{49} (-1)^k binom{99}{2k} = frac{(1 + i)^{99} + (1 - i)^{99}}{2}]We convert (1 + i) and (1 - i) to polar form. Both have a modulus of (sqrt{2}) and arguments (pi/4) and (-pi/4) respectively. Raising these to the 99th power:[(1 + i)^{99} = (sqrt{2})^{99} e^{i cdot 99 cdot pi/4} quad text{and} quad (1 - i)^{99} = (sqrt{2})^{99} e^{-i cdot 99 cdot pi/4}]Simplifying the angles modulo (2pi), we find:[99 cdot pi/4 equiv 3pi/4 pmod{2pi}]Thus, the expressions become:[(1 + i)^{99} = (sqrt{2})^{99} left( cos(3pi/4) + i sin(3pi/4) right) quad text{and} quad (1 - i)^{99} = (sqrt{2})^{99} left( cos(3pi/4) - i sin(3pi/4) right)]Adding these and dividing by 2:[frac{(1 + i)^{99} + (1 - i)^{99}}{2} = (sqrt{2})^{99} cos(3pi/4)]Since (cos(3pi/4) = -sqrt{2}/2), we have:[(sqrt{2})^{99} left( -frac{sqrt{2}}{2} right) = -(sqrt{2})^{100} cdot frac{1}{2} = -2^{50} cdot frac{1}{2} = -2^{49}]Thus, the sum from (k=0) to (49) is (-2^{49}). Subtracting the term for (k=0) (which is 1) gives the sum from (k=1) to (49):[-2^{49} - 1]However, since the answer choices do not include (-2^{49} - 1), we re-evaluate the problem and realize there might be a typo in the summation limits. The correct sum from (k=0) to (49) is (-2^{49}), which corresponds to option B.[boxed{B}]

🔑:1. Consider the binomial expansion of (1 + i)^{99}: [ (1 + i)^{99} = sum_{r=0}^{99} binom{99}{r} i^r ]2. Notice that the terms involving i alternate between the real part and the imaginary part because i^2 = -1.3. Specifically, the real part of the expression comes from the even powers of i. Therefore, we write: [ text{Real Part} = sum_{k=0}^{49} binom{99}{2k} (-1)^k ]4. Next, we can simplify (1 + i)^{99} algebraically using De Moivre's Theorem. Notice: [ (1+i)^{99} = left( sqrt{2} e^{i pi/4} right)^{99} = 2^{49.5} e^{99 i pi/4} ] [ = 2^{49.5} left( e^{25 i pi + 24.75 i pi/4} right) ] Recall that e^{2 pi i k} = 1 for any integer k, hence: [ = 2^{49.5} e^{24.75 i pi/4} ]5. Simplifying the angle, we get: [ 99pi/4 = 24pi + 3pi/4 rightarrow e^{i(99 pi / 4)} = e^{i 3 pi / 4} = -frac{1}{sqrt{2}} + i cdot frac{1}{sqrt{2}} ] 6. Therefore, the expanded form is: [ (1 + i)^{99} = 2^{49.5} left( -frac{1}{sqrt{2}} + i cdot frac{1}{sqrt{2}} right) = -2^{49} + 2^{49} i ]7. The real part of (1 + i)^{99} is: [ text{Real Part} = -2^{49} ]8. Comparing both results for the sum of the binomial coefficients and the real part of the expanded expression: [ sum_{k=0}^{49} binom{99}{2k} (-1)^k = -2^{49} ] Conclusion:The value of the sum is boxed{text{B}}

🧠:Alright, let me try to figure out how to approach this problem. So, the equation given is ( x^{2} + xy + y^{2} + x + y - 5 = 0 ), and we're told that one solution is ( x = 1 ), ( y = 1 ). I need to find other solutions maybe? Or perhaps verify that (1,1) is indeed a solution and explore the equation further. Let me start by checking if (1,1) actually satisfies the equation. Plugging in x=1 and y=1: ( 1^2 + (1)(1) + 1^2 + 1 + 1 -5 ). That's 1 + 1 + 1 + 1 + 1 -5 = 5 -5 = 0. Okay, so it checks out. Good. Now, maybe the question is to find all solutions? The problem statement isn't entirely clear, but given that it's a quadratic equation in two variables, there might be multiple solutions, and perhaps (1,1) is one of them. So, maybe the task is to find all real solutions?Let me think. The equation is a quadratic in both x and y. Such equations can represent conic sections. Let me recall that the general form of a quadratic equation in two variables is ( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 ). Comparing with our equation: A=1, B=1, C=1, D=1, E=1, F=-5. To classify the conic, we can compute the discriminant ( B^2 - 4AC ). Here, that would be ( 1 - 4(1)(1) = 1 - 4 = -3 ). Since the discriminant is negative, this is an ellipse. However, since it's a quadratic equation, an ellipse would mean that the set of solutions is a closed curve, so there are infinitely many solutions. Wait, but if it's an ellipse, then there are infinitely many real points on it. But maybe the problem is expecting integer solutions or something else? The given solution (1,1) is integer, so maybe we need to find all integer solutions? The original problem didn't specify, but since it was given a solution, maybe that's part of a system or there's more context. But let's assume that we need to find all real solutions unless specified otherwise.Wait, but the problem statement just gives the equation and mentions that (1,1) is a solution. Maybe the task is to find other solutions? The user hasn't specified exactly what they need, but since they mentioned "given one solution", perhaps they want to find all solutions or explain how to find them. Let's proceed under the assumption that we need to find all real solutions to the equation.To solve such an equation, we can try different methods: substitution, completing the square, rotation of axes to eliminate the xy term, etc. Let's try substitution first. Let me attempt to solve for one variable in terms of the other.Let me try solving for x in terms of y. The equation is quadratic in x: ( x^2 + (y + 1)x + (y^2 + y - 5) = 0 ). Using the quadratic formula, x would be ( [-(y + 1) pm sqrt{(y + 1)^2 - 4 cdot 1 cdot (y^2 + y -5)}]/2 ).Calculating the discriminant: ( (y + 1)^2 - 4(y^2 + y -5) = y^2 + 2y + 1 -4y^2 -4y +20 = -3y^2 -2y +21 ).So, for real solutions, the discriminant must be non-negative: ( -3y^2 -2y +21 geq 0 ). Multiply both sides by -1 (which reverses the inequality): ( 3y^2 + 2y -21 leq 0 ).Solving ( 3y^2 + 2y -21 =0 ). The discriminant here is ( 4 + 252 = 256 ). So y = [-2 ±16]/6. Thus, y = (14)/6 = 7/3 ≈2.333 or y = (-18)/6 = -3. So the inequality ( 3y^2 + 2y -21 leq 0 ) holds between y=-3 and y=7/3. Therefore, real solutions exist only when y is in [-3, 7/3]. Similarly, we could do the same for x if we solve for y in terms of x.But this seems a bit messy. Alternatively, maybe we can use symmetry or substitution. Let me check if the equation is symmetric in x and y. Looking at the equation: ( x^2 + xy + y^2 + x + y -5 =0 ). Yes, swapping x and y gives the same equation. So the equation is symmetric in x and y. Therefore, if (a,b) is a solution, then (b,a) is also a solution. So, since (1,1) is a solution, that's on the line y=x. Other solutions may come in pairs where x and y are swapped.This symmetry might be useful. Maybe we can set x = y to find solutions on the line y=x. Let's try that. If x = y, substitute into the equation:( x^2 + x^2 + x^2 + x + x -5 = 0 Rightarrow 3x^2 + 2x -5 =0 ).Solving this quadratic equation: discriminant = 4 + 60 = 64. So x = [-2 ±8]/6. Thus, x = (6)/6=1 or x = (-10)/6=-5/3. Therefore, when x=y, the solutions are (1,1) and (-5/3, -5/3). So we've found another solution: (-5/3, -5/3). Let me verify this. Plugging x=y=-5/3:Left side: (-5/3)^2 + (-5/3)(-5/3) + (-5/3)^2 + (-5/3) + (-5/3) -5Calculating each term:First term: 25/9Second term: 25/9Third term: 25/9Fourth term: -5/3Fifth term: -5/3Last term: -5Adding them up:25/9 +25/9 +25/9 = 75/9 = 25/3Then, -5/3 -5/3 = -10/3Then, -5 = -15/3Total: 25/3 -10/3 -15/3 = (25 -10 -15)/3 = 0/3 =0. Correct. So (-5/3, -5/3) is another solution.So, so far, we have two solutions on the line y=x: (1,1) and (-5/3, -5/3). But there might be more solutions where x ≠ y.To find all solutions, perhaps we can parametrize the equation. Alternatively, use substitution. Let me try subtracting the equation from itself in a clever way. Wait, perhaps we can use a substitution like u = x + y and v = x - y. Sometimes rotating the coordinates helps. Since the equation is quadratic and has an xy term, maybe rotating the axes to eliminate the xy term would simplify it.The general method for rotating conic sections to eliminate the xy term involves rotating the axes by an angle θ where tan(2θ) = B/(A - C). In our case, A=1, B=1, C=1. So tan(2θ) = 1/(1 -1) = 1/0, which is undefined. Therefore, 2θ = π/2, so θ = π/4. That is, rotating by 45 degrees.So let's perform a rotation by θ=45°. The rotation formulas are:x = x'cosθ - y'sinθy = x'sinθ + y'cosθSince θ=45°, cosθ = sinθ = √2/2. Therefore:x = (x' - y')√2/2y = (x' + y')√2/2Now, substitute these into the original equation:First, compute x^2 + xy + y^2 +x + y -5 =0.Let me compute each term step by step.Compute x^2:[(x' - y')√2/2]^2 = (x'^2 - 2x'y' + y'^2)(2/4) = (x'^2 - 2x'y' + y'^2)(1/2)Compute y^2:[(x' + y')√2/2]^2 = (x'^2 + 2x'y' + y'^2)(1/2)Compute xy:[(x' - y')√2/2][(x' + y')√2/2] = [(x'^2 - y'^2) * 2/4] = (x'^2 - y'^2)/2Now, sum x^2 + xy + y^2:= [ (x'^2 - 2x'y' + y'^2)/2 ] + [ (x'^2 - y'^2)/2 ] + [ (x'^2 + 2x'y' + y'^2)/2 ]Combine the terms:First term: (x'^2 - 2x'y' + y'^2)/2Second term: (x'^2 - y'^2)/2Third term: (x'^2 + 2x'y' + y'^2)/2Adding them together:Numerator:(x'^2 -2x'y' + y'^2) + (x'^2 - y'^2) + (x'^2 +2x'y' + y'^2)Combine like terms:x'^2 terms: 1 +1 +1 =3x'^2y'^2 terms:1 -1 +1=1y'^2Cross terms: -2x'y' +0 +2x'y'=0So total numerator: 3x'^2 + y'^2Therefore, x^2 + xy + y^2 = (3x'^2 + y'^2)/2Now, compute the linear terms x + y:x + y = [(x' - y')√2/2] + [(x' + y')√2/2] = [x' - y' + x' + y']√2/2 = (2x')√2/2 = x'√2Therefore, the entire equation becomes:(3x'^2 + y'^2)/2 + x'√2 -5 =0Multiply both sides by 2 to eliminate the denominator:3x'^2 + y'^2 + 2x'√2 -10 =0Now, let's rearrange terms:3x'^2 + 2x'√2 + y'^2 =10To complete the square for the x' terms:3x'^2 + 2x'√2 = 3(x'^2 + (2√2/3)x') Complete the square inside the parentheses:x'^2 + (2√2/3)x' = x'^2 + (2√2/3)x' + ( (√2/3)^2 ) - ( (√2/3)^2 )= [x' + √2/3]^2 - (2/9)Multiply back by 3:3[ x'^2 + (2√2/3)x' ] = 3[ (x' + √2/3)^2 - 2/9 ] = 3(x' + √2/3)^2 - 2/3Therefore, substituting back into the equation:3(x' + √2/3)^2 -2/3 + y'^2 =10Bring constants to the right:3(x' + √2/3)^2 + y'^2 =10 + 2/3 =32/3Divide both sides by 32/3 to get the standard form:[3(x' + √2/3)^2]/(32/3) + y'^2/(32/3) =1Simplify:(x' + √2/3)^2/(32/9) + y'^2/(32/3) =1Which can be written as:(x' + √2/3)^2/( (4√2)/3 )^2 + y'^2/( (4√6)/3 )^2 =1This is the equation of an ellipse centered at (x', y') = (-√2/3, 0) in the rotated coordinate system, with semi-major axis (4√6)/3 along the y' axis and semi-minor axis (4√2)/3 along the x' axis.So, this confirms that the original equation represents an ellipse, and thus has infinitely many real solutions. However, since we were given one solution (1,1), maybe the problem is expecting us to find integer solutions? Let's check if there are more integer solutions.Let me check integer values around (1,1). Let's see.Suppose x and y are integers. Then the equation is ( x^2 + xy + y^2 + x + y =5 ). Let's try small integers.Test x=0:0 +0 + y^2 +0 + y = y² + y =5. So y² + y -5=0. Discriminant:1 +20=21. Not a square. So no integer solutions.x=1:1 + y + y² +1 + y = y² +2y +2 =5 → y² +2y -3=0. Solutions y=(-2 ±√(4 +12))/2= (-2 ±4)/2 → y=1 or y=-3. So when x=1, y=1 or y=-3. So (1,1) is given, another solution is (1, -3). Similarly, due to symmetry, (-3,1) is also a solution.Wait, let me check x=1, y=-3:Plug into equation: 1 + (1)(-3) + (-3)^2 +1 + (-3) -5 =1 -3 +9 +1 -3 -5= (1 -3) + (9 +1) + (-3 -5)= (-2) +10 + (-8)=0. Correct. So (1,-3) is a solution. Similarly, (-3,1) is a solution.x=2:4 +2y + y² +2 + y = y² +3y +6=5 → y² +3y +1=0. Discriminant 9 -4=5. Not integer.x=-1:1 + (-1)y + y² -1 + y = y² +0y +0=5 → y²=5. Not integer.x=-2:4 + (-2)y + y² -2 + y = y² - y +2=5 → y² - y -3=0. Discriminant 1 +12=13. Not square.x=3:9 +3y + y² +3 + y = y² +4y +12=5 → y² +4y +7=0. Discriminant 16-28=-12. No real solutions.x=-3:9 + (-3)y + y² -3 + y = y² -2y +6=5 → y² -2y +1=0 → (y-1)^2=0 → y=1. So (-3,1) is a solution, which we already have.Similarly, y=1, solving for x:x² +x(1) +1 +x +1 -5 =x² +2x +2 -5 =x² +2x -3=0. Solutions x=(-2 ±√(4 +12))/2= (-2 ±4)/2=1 or -3. So (1,1) and (-3,1). Which matches.Similarly, y=-3:x² +x(-3) +9 +x -3 -5 =x² -2x +1=0 → (x-1)^2=0 → x=1. So (1,-3) is a solution.Therefore, the integer solutions are (1,1), (1,-3), (-3,1), and (-5/3,-5/3) is a non-integer solution. Wait, but earlier we found (-5/3,-5/3) as a solution when x=y, but that's not integer. So, the integer solutions are the three points: (1,1), (1,-3), (-3,1). Wait, but (1,1) is on y=x, and the other two are symmetric across y=x.Wait, but when x=1, y=-3 and when y=1, x=-3. So those are two distinct points. So total of three integer solutions: (1,1), (1,-3), (-3,1). Wait, but (1,-3) and (-3,1) are distinct.Let me check if there are more integer solutions. Let's try y=2:Original equation: x² +2x +4 +x +2 -5 =x² +3x +1=0. Discriminant 9 -4=5. Not integer.y=-1:x² -x +1 +x -1 -5 =x² -0x -5=0. x²=5. No integer solutions.y=-2:x² -2x +4 +x -2 -5 =x² -x -3=0. Discriminant 1 +12=13. Not square.So no other integer solutions. Therefore, the integer solutions are (1,1), (1,-3), (-3,1). But we also have the non-integer solution (-5/3,-5/3). But since the equation is an ellipse, there are infinitely many real solutions, but only three integer solutions.But the original problem statement only mentions that one solution is (1,1). Depending on the context, maybe the user wants all solutions, or perhaps just the integer ones. But since the problem didn't specify, but mentioned that (1,1) is given, perhaps the task is to find other solutions or confirm that it's the only integer solution. However, we found three integer solutions. Wait, but (1,1) is given, and we found (1,-3) and (-3,1) as others. So perhaps the answer is to list all solutions, but since it's an ellipse, there are infinitely many. However, if the problem is from a context where integer solutions are expected (like a Diophantine equation), then the integer solutions are the three mentioned.But the problem statement just says "given one solution: x=1, y=1". It doesn't specify the task. Maybe the task is to find other solutions? Or perhaps to solve the system where this equation is paired with another, but no, only one equation is given. Wait, one equation with two variables usually has infinitely many solutions. But maybe the user wants to express y in terms of x or vice versa, or find particular solutions. Since (1,1) is given, perhaps the problem is part of a system, but since only one equation is presented, perhaps the user just wants to verify that (1,1) is a solution and maybe find another solution.Alternatively, maybe the original problem was to find the solution(s) given that x=1 or y=1, but since it's presented as given, perhaps the user expects more. Alternatively, maybe the problem is to find the solution closest to the origin or something else.Alternatively, perhaps the problem is part of a system where another equation is missing. But as per the current information, we have one equation with two variables, hence infinitely many solutions. But the user provided that one solution is (1,1), perhaps implying that there are more solutions to find, but without additional constraints, we can't determine a unique solution.Alternatively, maybe the user wants to find all solutions where x and y are real numbers, which we've characterized as an ellipse, but perhaps parametrize it.Alternatively, to find the intersection points with particular lines, like y=x, which we did, giving (1,1) and (-5/3,-5/3). And the other integer solutions by plugging in x=1 or y=1.Alternatively, maybe the user is expecting to use the given solution to factor the equation. Let's see. If (x=1, y=1) is a solution, perhaps the equation can be factored as (x -1)(something) + (y -1)(something) =0, but factoring multivariate polynomials is more complex.Alternatively, we can write the equation as ( x^2 + xy + y^2 + x + y -5 =0 ). Let me try to rearrange terms:( x^2 + xy + y^2 + x + y =5 ).Notice that ( x^2 + xy + y^2 = frac{3}{4}(x + y)^2 + frac{1}{4}(x - y)^2 ), but not sure if that helps. Alternatively, completing the square.Let me group terms:( x^2 + x + xy + y^2 + y =5 ).Group x terms and y terms:( x^2 + x + xy + y^2 + y = x(x +1) + y(y +1) + xy =5 ). Not sure.Alternatively, consider adding and subtracting terms to complete squares.Let me try:( x^2 + xy + y^2 = (x + y/2)^2 + (3/4)y^2 ). This is a technique used to complete the square for expressions with xy terms.So, ( x^2 + xy + y^2 = (x + y/2)^2 + (3/4)y^2 ).Therefore, the equation becomes:( (x + y/2)^2 + (3/4)y^2 + x + y -5 =0 ).Hmm, not sure if this helps. Alternatively, let's consider variables substitution. Let u = x + y and v = x - y. Then, x = (u + v)/2 and y = (u - v)/2. Substitute into the equation:First, compute x^2 + xy + y^2:x^2 + xy + y^2 = [(u + v)/2]^2 + [(u + v)/2][(u - v)/2] + [(u - v)/2]^2Compute each term:First term: (u² + 2uv + v²)/4Second term: (u² - v²)/4Third term: (u² - 2uv + v²)/4Sum them up:[(u² +2uv +v²) + (u² -v²) + (u² -2uv +v²)] /4= [3u² +0uv + v²]/4 = (3u² + v²)/4Then, the linear terms x + y = u.Therefore, the equation becomes:(3u² + v²)/4 + u -5 =0 → multiply by 4:3u² + v² +4u -20 =0Rearrange:3u² +4u + v² =20Complete the square for u:3(u² + (4/3)u) + v² =20= 3[ u² + (4/3)u + (4/9) ] - 3*(4/9) + v² =20= 3(u + 2/3)^2 - 4/3 + v² =20→ 3(u + 2/3)^2 + v² =64/3Divide by 64/3:[3(u + 2/3)^2]/(64/3) + v²/(64/3) =1Simplify:(u + 2/3)^2/(64/9) + v²/(64/3)=1Which is an ellipse in terms of u and v. So similar to the previous rotated form, but different parameters. This might not be particularly helpful, but shows the equation's elliptical nature.Alternatively, perhaps using parametric equations. For an ellipse, we can parametrize using trigonometric functions. However, without specific requirements, it's hard to know what form is desired.Given that the problem mentions a systematic thinking process, but the final answer should be precise and accurate. Since the user hasn't specified what exactly is required, but given that (1,1) is a solution, and through exploration, we found other solutions, both integer and non-integer. However, considering the initial problem statement, which just provides the equation and mentions (1,1) is a solution, the likely expected answer is to find other solutions, possibly integer ones. Since we found (1,-3) and (-3,1) as integer solutions, and (-5/3,-5/3) as another real solution, those could be mentioned.Alternatively, if the task is to find all solutions, but since it's an ellipse, they form a continuous set. However, without additional constraints, we can't list them all. Therefore, the answer probably expects mentioning the other integer solutions and perhaps the other intersection with y=x.But let's check the original problem statement again: "given one solution: x=1; y=1". The instruction was in Chinese, translated to English, asking to explore the problem with a long thinking process before providing the solution. The user might want to know all real solutions, but since that's an ellipse, it's a bit involved. Alternatively, they might want to know the integer solutions. Since we found three integer solutions: (1,1), (1,-3), (-3,1), and the other solution on y=x being (-5/3,-5/3).Alternatively, if we consider that the original problem might have been part of a system where another equation is involved, but only one is given, leading to infinitely many solutions. However, given the information, the best we can do is list the solutions we found.But to confirm, let's check if there are more solutions where x and y are both integers beyond what we already found. Trying y=4:Plugging y=4 into the equation: x² +4x +16 +x +4 -5 =x² +5x +15=0. Discriminant 25-60=-35. No real solutions.y=-4:x² -4x +16 +x -4 -5 =x² -3x +7=0. Discriminant 9-28=-19. No real solutions.So no more integer solutions. Therefore, the integer solutions are (1,1), (1,-3), (-3,1). The other real solution on the line y=x is (-5/3,-5/3). So, these are specific solutions.Alternatively, if the task is to express y in terms of x or vice versa, the earlier steps using the quadratic formula can be used. For example, solving for y in terms of x:Starting with the original equation: ( x^2 + xy + y^2 +x + y -5 =0 ). Rearranged as y² + (x +1)y + (x² +x -5)=0.Using quadratic formula for y:y = [-(x +1) ± sqrt((x +1)^2 -4*1*(x² +x -5))]/2Calculate discriminant:( (x +1)^2 -4(x² +x -5) =x² +2x +1 -4x² -4x +20= -3x² -2x +21 ).Thus, y = [ -x -1 ± sqrt(-3x² -2x +21) ] /2.For real solutions, the discriminant must be non-negative:-3x² -2x +21 ≥0 →3x² +2x -21 ≤0.Solving 3x² +2x -21=0, we found roots at x=(-1 ±√64)/3=(-1 ±8)/3. Thus, x=7/3 or x=-3. So the inequality holds for x between -3 and 7/3. Therefore, real solutions exist for x ∈ [-3, 7/3]. Similarly for y.Thus, the solutions are pairs (x,y) where x ∈ [-3,7/3] and y is given by the quadratic formula above, leading to two possible y for each x in that interval, except possibly at the endpoints where discriminant is zero.At x=7/3, discriminant is zero:y = [ -7/3 -1 ±0 ] /2 = (-10/3)/2 = -5/3. So at x=7/3, y=-5/3. But wait, when x=7/3, plugging into the original equation, we should get y=-5/3. But earlier, when we set x=y, we found (-5/3,-5/3). Wait, this seems contradictory.Wait, let me check. If x=7/3, then the discriminant is zero, so y = [ -7/3 -1 ] /2 = (-10/3)/2 = -5/3. So the point (7/3, -5/3) is a solution. Similarly, at x=-3, discriminant is zero:y = [ -(-3) -1 ±0 ] /2 = (3 -1)/2 =1. So y=1. So the point (-3,1) is a solution. Which matches our earlier integer solution.Similarly, at x=1, discriminant sqrt(-3(1)^2 -2(1)+21)=sqrt(-3-2+21)=sqrt(16)=4. So y=[-1-1 ±4]/2= (-2 ±4)/2 → y=1 or y=-3. Which matches our earlier result.Therefore, parametrizing all solutions, we can write them as:For x ∈ [-3,7/3], y = [ -x -1 ± sqrt(-3x² -2x +21) ] /2.Similarly, we can write x in terms of y.In conclusion, the equation represents an ellipse with infinitely many real solutions, three of which are integer points: (1,1), (1,-3), (-3,1), and another notable solution at (-5/3,-5/3) where x=y.Given the problem's context and the information provided, the most precise answer likely includes all integer solutions and possibly the other notable solution on the line y=x. However, since the user mentioned providing the final precise and accurate solution, and considering that the original problem statement mentioned only one solution, the comprehensive answer should include all solutions, but given the nature of the equation, specifying all might not be feasible. Therefore, the key solutions are the integer ones and the other intersection with y=x.But wait, the problem didn't specify whether it wants real solutions or integer solutions. Since (1,1) is an integer solution, but the equation is over real numbers, it's ambiguous. However, given the systematic thinking process, it's appropriate to mention both the integer solutions and the nature of the equation as an ellipse with infinitely many real solutions.But since the user instruction says "provide the final precise and accurate solutions", and given that in many math problems, especially when a solution is given as an integer, finding all integer solutions is expected. Therefore, the final answer should list all integer solutions: (1,1), (1,-3), (-3,1). However, it's also possible that the problem expects the other solution on the line y=x, which is (-5/3,-5/3), so that could be included as a non-integer solution.Alternatively, if the problem is part of a system where x and y are required to satisfy another condition, but since only one equation is given, we can't determine. Given the information, the most comprehensive precise answer is to list all integer solutions and note the other real solution on y=x.But to verify again, let's check if there are more solutions when substituting different values.For example, take y=0:Equation becomes x² +0 +0 +x +0 -5=0 →x² +x -5=0. Solutions x=(-1 ±√21)/2. Not integer.y=2:x² +2x +4 +x +2 -5 =x² +3x +1=0. Solutions x=(-3 ±√5)/2. Not integer.y=-2:x² -2x +4 +x -2 -5 =x² -x -3=0. Solutions x=(1 ±√13)/2. Not integer.Thus, confirming no other integer solutions.Therefore, the integer solutions are (1,1), (1,-3), (-3,1). The other real solution on the line y=x is (-5/3,-5/3). If non-integer solutions are acceptable, then there's an infinite number, but these are specific ones.But since the problem mentioned "given one solution", perhaps the task is to find all solutions, but given the open-ended nature, the answer should probably include the integer solutions and the other intersection point.However, depending on the user's intention, the answer might vary. Given that in many Olympiad problems or algebra problems, finding integer solutions is common, especially when an integer solution is provided. Therefore, the final answer is likely the three integer solutions.

✅:The given equation is ( x^2 + xy + y^2 + x + y - 5 = 0 ), with one solution provided: ( (1, 1) ). Here's the systematic exploration and solutions: Key Steps:1. Verification: Confirmed ( (1, 1) ) satisfies the equation.2. Conic Classification: The equation represents an ellipse (discriminant ( B^2 - 4AC = -3 )).3. Symmetry: The equation is symmetric in ( x ) and ( y ), implying solutions like ( (a, b) ) and ( (b, a) ).4. Integer Solutions: - Tested integer values, leading to: - ( (1, -3) ) and ( (-3, 1) ) via substitution.5. Line ( y = x ): - Found ( (-5/3, -5/3) ) by solving ( 3x^2 + 2x - 5 = 0 ). Solutions:- Integer Solutions: - ( boxed{(1, 1)} ) - ( boxed{(1, -3)} ) - ( boxed{(-3, 1)} )- Non-Integer Solution on ( y = x ): - ( boxed{left(-dfrac{5}{3}, -dfrac{5}{3}right)} ) Conclusion:The equation has infinitely many real solutions forming an ellipse, with three integer solutions and a notable non-integer solution on the line ( y = x ).

🔑:Given the equation (x^2 + xy + y^2 + x + y - 5 = 0) and one solution (x = 1), (y = 1), we will attempt to find all possible solutions.1. Substitute (x = 1) and (y = 1) to verify the solution: [ 1^2 + 1 cdot 1 + 1^2 + 1 + 1 - 5 = 1 + 1 + 1 + 1 + 1 - 5 = 0 ] This confirms that (x = 1) and (y = 1) is indeed a solution.2. Change of variables: Let (x = x_1 + 1) and (y = y_1 + 1). Substituting these into the original equation gives: [ (x_1 + 1)^2 + (x_1 + 1)(y_1 + 1) + (y_1 + 1)^2 + (x_1 + 1) + (y_1 + 1) - 5 = 0 ]3. Expand and simplify: [ x_1^2 + 2x_1 + 1 + x_1 y_1 + x_1 + y_1 + 1 + y_1^2 + 2y_1 + 1 + x_1 + 1 + y_1 + 1 - 5 = 0 ] [ x_1^2 + y_1^2 + x_1 y_1 + 4x_1 + 4y_1 = 0 ]4. Attempt substitution (x_1 = 0) and (y_1 = 0): [ 0^2 + 0^2 + 0 cdot 0 + 4 cdot 0 + 4 cdot 0 = 0 ] This is trivially true, providing one known solution: [ (x_1, y_1) = (0, 0) implies (x, y) = (1, 1) ]5. Consider the case ( frac{x_1}{y_1} = frac{m}{n} ) where (m) and (n) are coprime integers: [ x_1 = km quad text{and} quad y_1 = kn quad text{for some integer } k ] Substitute (x_1) and (y_1): [ k^2m^2 + k^2n^2 + k^2mn + 4km + 4kn = 0 ] Divide by (k^2): [ m^2 + n^2 + mn + frac{4m}{k} + frac{4n}{k} = 0 ]6. Express (k): [ k = -frac{4m(m+n)}{m^2+mn+n^2} ] [ k = -frac{4n(m+n)}{m^2+mn+n^2} ]7. Consider (m^2 + mn + n^2): - Since (m) and (n) are coprime, (m(m+n)) and (m^2 + mn + n^2) are also coprime. - Therefore, (m^2 + mn + n^2) must divide (4). Possible cases are when (m^2 + mn + n^2 = 1, 2, text{ or } 4).8. Case by case analysis: Case 1: (m^2 + mn + n^2 = 1) [ m = pm 1, n = 0 implies x_1 = -4, y_1 = 0 implies (x, y) = (-3, 1) ] Case 2: (m^2 + mn + n^2 = 2) [ m = frac{-n pm sqrt{8-3n^2}}{2}, |n| leq sqrt{frac{8}{3}}, |n| = 1 ] - No integer solutions for (m) (cannot form a perfect square inside the square root). Case 3: (m^2 + mn + n^2 = 4) [ |n| leq sqrt{frac{16}{3}}, |n| = 1 text{ or } 2 ] For (n = pm 2): [ x = 1, y = -3 text { or } y = 1, x = -3 implies (1, -3), (-3, 1) ] For (|n| = 1): - (m) will be irrational, hence discarded. Conclusion: The solutions to the equation are: [ (1, 1), (1, -3), (-3, 1) ] (boxed{(1,1),(1,-3),(-3,1)})

🧠:Okay, let's see. The problem is asking to find all bijective functions (which means permutations) f of the set {x₁, x₂, ..., xₙ} where n is an odd number ≥3, such that the absolute differences |f(xᵢ) - xᵢ| are the same for all i from 1 to n. So, all elements are moved by the same distance when we apply the permutation f. Hmm, interesting.First, let me rephrase the problem in my own words. We have a set of real numbers, and we need to permute them such that each number is shifted by the same absolute amount. But since it's a permutation, every element has to go to a unique position. So, for example, if we have numbers arranged on the number line, each number has to move either to the right or left by exactly some fixed distance d. But since it's a permutation, the positions after shifting must also be the original set. So, shifting all elements by a fixed distance d would only work if the entire set is symmetric with respect to some point, right? Because otherwise, shifting would take some elements outside the original set.Wait, but in this problem, the set {x₁, x₂, ..., xₙ} is arbitrary. The problem states that the xᵢ are real numbers, but doesn't specify they are in any particular order or have any particular structure. So, maybe the functions f have to work for any such set? Or is the set given, and we have to find f for that specific set? Hmm, the problem says "Find all bijective functions...", so I think it's for any such set with n elements where n is odd and ≥3. Wait, no, maybe the set is fixed, and we need to find the bijections on that set satisfying the condition. The problem statement is a bit ambiguous. Let me check again.The problem says: Let n be an odd number not less than 3, and x₁, x₂, ..., xₙ be real numbers. Find all bijective functions f: {x₁, x₂, ..., xₙ} → {x₁, x₂, ..., xₙ} such that |f(x₁) - x₁| = ... = |f(xₙ) - xₙ|. So, the set is given, but the xᵢ are arbitrary real numbers. So, the functions f must satisfy that condition for that specific set. Therefore, depending on how the xᵢ are arranged, the possible permutations f would be different.But the problem is asking for all such bijections, so perhaps the answer is that f must be a composition of a translation (shift by d) and a reflection if necessary, but since the set is finite, shifting all elements by d may not map the set onto itself unless the set is symmetric. Wait, but the set is arbitrary. So maybe the only way this works is if d = 0, which would mean f is the identity permutation. But wait, that's only if shifting by d is not possible. Alternatively, maybe there's another structure.Wait, but the problem allows the permutation to move each element by the same absolute difference, but not necessarily the same direction. So, for each element xᵢ, f(xᵢ) can be either xᵢ + d or xᵢ - d. However, since the permutation is bijective, each element must be mapped to a unique element. So, for example, if I have elements x₁, x₂, x₃, and I decide that each element is shifted by d, but then x₁ + d has to be another element in the set, and similarly for x₂ + d and x₃ + d, but since it's a permutation, these images must all be distinct. So, the set must be arranged such that adding or subtracting d maps each element to another element in the set.But since the set is arbitrary, unless the set has some kind of arithmetic progression or symmetry, such permutations might not exist. However, the problem states that n is odd. Hmm. Maybe the key is that for an odd-sized set, such a permutation can only be the identity? Because if you try to shift all elements by a non-zero d, you need the set to be symmetric, but with an odd number of elements, there's a central element which would have nowhere to go if you shift by d. Let's consider an example.Suppose n=3, and the elements are x₁, x₂, x₃ arranged in increasing order. Suppose we try to shift each element by d. If the set is symmetric around some point, say x₂ is the center. Then shifting left by d would map x₁ to x₂ - d, x₂ to x₂, x₃ to x₂ + d. Wait, but unless the set is symmetric, such shifts would not map to elements in the set. Alternatively, if the set is an arithmetic progression. Suppose x₁ = a - h, x₂ = a, x₃ = a + h. Then shifting by h would map x₁ to a, x₂ to a + h, x₃ to a + 2h, but a + 2h is not in the set. Unless h = 0, which would be the identity. Alternatively, shifting by 2h would wrap around, but in this case, the set is not closed under addition.Alternatively, maybe each element is paired with another element at distance d, but since n is odd, one element would be left unpaired, which has to map to itself. So, if d ≠ 0, then we have (n-1)/2 pairs of elements each separated by 2d, and the middle element is fixed. Wait, that might be possible.For example, let's take n=3. Suppose we have elements x, y, z where y - x = z - y = d. Then the permutation f could swap x and z, and keep y fixed. Then |f(x) - x| = |z - x| = 2d, |f(z) - z| = |x - z| = 2d, and |f(y) - y| = 0. But the problem requires all |f(xᵢ) - xᵢ| to be equal. In this case, two have 2d and one has 0, which is not equal unless d=0. So, that doesn't work. Hmm.Alternatively, maybe if the set is arranged such that each element is d apart from another element, but with an odd number, one element remains. Wait, but if all |f(xᵢ) - xᵢ| = d, then the permutation would have to consist of transpositions (swaps) of elements that are d apart, but if n is odd, there must be at least one fixed point. However, the fixed point would have |f(x) - x| = 0, which must equal d, so d must be 0. Therefore, the only possible permutation is the identity function. Is this the case?Let me formalize this. Suppose there exists a permutation f such that |f(xᵢ) - xᵢ| = d for all i. If d ≠ 0, then for each xᵢ, f(xᵢ) must be xᵢ ± d. Since f is a bijection, each image must be unique. So, we need that for each xᵢ, either xᵢ + d or xᵢ - d is in the set, and all these images are distinct. However, since the set has an odd number of elements, if we pair each xᵢ with xᵢ + d or xᵢ - d, there must be one element left unpaired, which would have to map to itself. But then |f(x) - x| = 0 for that element, which would imply d = 0. Hence, the only possibility is d = 0, so f is the identity permutation.Therefore, the conclusion is that the only such bijection is the identity function. Let me check this with an example. Take n=3, elements 0, 1, 2. Suppose we try to have each element moved by 1. Then 0 can go to 1, 1 can go to 0 or 2, and 2 can go to 1. But if 0→1 and 1→0, then 2 must map to itself, which gives |2 - 2| = 0, which doesn't equal 1. Alternatively, 0→1, 1→2, 2→0? Then the differences are |1-0|=1, |2-1|=1, |0-2|=2, which is not equal. So that doesn't work. If we try to shift all by 1, but since the set isn't circular, it's impossible. Hence, only identity works.Another example: n=5, elements -2, -1, 0, 1, 2. Suppose we try to shift each by 1. Then -2→-1, -1→0, 0→1, 1→2, 2→? But 2 +1=3 is not in the set. Alternatively, shift left by 1: -2→-3 (not in set). So shifting doesn't work. Alternatively, swap -2 and -1, 0 stays, 1 and 2 swap. Then differences are 1, 1, 0, 1, 1. Not equal. Hence, only identity has all differences zero.Therefore, the only possible bijection is the identity function. Therefore, the answer is that the only such function is the identity function.Wait, but let me think again. Suppose the set is symmetric around some point. For example, consider the set {-1, 0, 1}. If we define f(-1) = 1, f(1) = -1, and f(0) = 0. Then the differences are |1 - (-1)| = 2, | -1 -1| = 2, and |0 - 0| = 0. Not equal. So, again, fixed point causes a problem.Alternatively, if we have another set where each element is paired with another element at distance d, and the middle element is at distance d/2 from some other point? Not sure. For example, if the set is {0, d, 2d}, but n=3. Then swapping 0 and 2d, keeping d fixed. Then the differences are 2d, 0, 2d. Not equal. So again, no.Alternatively, maybe the set is arranged in such a way that each element can be shifted by d to another element. For example, consider a circular arrangement, but with real numbers, you can't wrap around. So that's not possible unless the set is finite and circular, which isn't the case here.Alternatively, maybe the permutation is such that it's a product of transpositions where each transposition swaps two elements at distance 2d apart, and the fixed point is shifted by d. Wait, but then the fixed point would have difference d, but others would have 2d. Not equal. So that's not possible.Wait, unless all elements are equally spaced. Let's suppose the set is an arithmetic progression: x₁, x₁ + h, x₁ + 2h, ..., x₁ + (n-1)h. Then, if we take d = kh for some integer k. Then shifting each element by kh would map them to another element, but for an arithmetic progression with n elements, shifting by kh modulo nh? But real numbers don't wrap around. Unless the set is arranged periodically, which isn't the case. So shifting would take elements out of the set unless k=0. Hence, only identity works.Alternatively, maybe reflecting the set around its midpoint. Suppose the set is symmetric around some central point. For example, if the set is {a, a + h, a + 2h, ..., a + (n-1)h} and n is odd, then the central element is a + ((n-1)/2)h. If we reflect each element around the central element, then each element xᵢ would map to 2a + (n-1)h - xᵢ. Then the difference |f(xᵢ) - xᵢ| would be |2a + (n-1)h - 2xᵢ|. For the set to be symmetric, 2a + (n-1)h - xᵢ must also be in the set. But unless the set is symmetric around the central element, which it is. For example, in the set {1, 2, 3, 4, 5}, reflecting around 3 would give f(1)=5, f(2)=4, f(3)=3, f(4)=2, f(5)=1. Then the differences are |5 -1|=4, |4 -2|=2, |3 -3|=0, |2 -4|=2, |1 -5|=4. These are not equal. Hence, even in symmetric sets, reflection does not yield equal differences. The differences depend on the distance from the center. So that approach doesn't work either.Another approach: suppose all elements are mapped to another element at the same distance d. Then, for each xᵢ, f(xᵢ) = xᵢ ± d. Since it's a permutation, we need that for every xᵢ, either xᵢ + d or xᵢ - d is present in the set, and all these images are distinct. However, with n being odd, we can't pair all elements into such pairs; one element must map to itself, forcing d = 0. Therefore, the only possible permutation is the identity.To confirm, suppose d ≠ 0. Then each element is part of a pair (x, x + d) or (x, x - d). Since the permutation is bijective, each element must be paired uniquely. But with n odd, there must be an unpaired element, say x₀, which must satisfy x₀ + d = x₀ or x₀ - d = x₀, which implies d = 0, contradicting d ≠ 0. Therefore, no such permutation exists unless d = 0, which corresponds to the identity function.Therefore, the only bijective function satisfying the condition is the identity function.

✅:The problem requires finding all bijective functions ( f ) on a set of ( n ) real numbers (where ( n ) is an odd integer ( geq 3 )) such that the absolute difference ( |f(x_i) - x_i| ) is constant for all ( i ). Key Steps in the Reasoning:1. Bijection and Constant Difference: For ( f ) to satisfy the condition, each element ( x_i ) must map to ( x_i pm d ), where ( d ) is a fixed constant. Since ( f ) is a bijection, every image ( f(x_i) ) must be unique and belong to the original set.2. Pairing Elements: If ( d neq 0 ), elements must form pairs ((x, x + d)) or ((x, x - d)). However, since ( n ) is odd, pairing all elements is impossible—one element must remain fixed. A fixed element implies ( |f(x) - x| = 0 ), forcing ( d = 0 ).3. Conclusion for ( d ): The only way to avoid contradiction (non-zero ( d ) with a fixed point) is ( d = 0 ). Thus, ( f(x_i) = x_i ) for all ( i ), meaning ( f ) is the identity function.Final Answer: The only such bijective function is the identity function. Therefore, the solution is [boxed{f(x_i) = x_i text{ for all } i}]

🔑:The given problem requires finding all bijective functions f:left{x_1, x_2, cdots, x_nright} rightarrow left{x_1, x_2, cdots, x_nright}such that left|f(x_i) - x_iright| = a quad text{for all } i = 1,2,cdots,n,where ( n ) is an odd number and ( x_1, x_2, cdots, x_n ) are real numbers.1. Identify Function Property: [ left|f(x_i) - x_iright| = a quad text{implies that} quad f(x_i) = x_i pm a quad text{for each } i. ]2. Sum Study: Note that ( f ) maps the set ({x_1, x_2, dots, x_n}) to itself. Therefore, it must hold that the sum of the values under ( f ) equals the sum of the original values: [ sum_{i=1}^{n} f(x_i) = sum_{i=1}^{n} x_i. ]3. Difference in Sums: Compute the difference of the sums: [ 0 = sum_{i=1}^{n} f(x_i) - sum_{i=1}^{n} x_i = sum_{i=1}^{n} left[f(x_i) - x_iright]. ]4. Form Expression: Using ( f(x_i) = x_i pm a ): [ sum_{i=1}^{n} left[f(x_i) - x_iright] = (x_1 pm a - x_1) + (x_2 pm a - x_2) + ldots + (x_n pm a - x_n). ]5. Simplify: Each term simplifies to ( pm a ): [ sum_{i=1}^{n} left[f(x_i) - x_iright] = pm a pm a pm a cdots pm a, ] where there are ( n ) terms. Let ( m ) be the number of ( +a ) terms and ( k ) be the number of ( -a ) terms such that ( m + k = n ). 6. Total Sum: This results in: [ sum_{i=1}^{n} left[f(x_i) - x_iright] = ma - ka = (m-k)a. ]7. Condition for Zero Sum: Since the sum must be zero: [ (m - k)a = 0. ]8. Odd Number Implication: Since ( n ) is odd, ( m ) and ( k ) cannot both be equal. Therefore, ( a ) must be zero for the equation to hold: [ (m - k)a = 0 implies a = 0 quad text{since } m neq k. ]9. Final Function: Thus, ( f(x_i) = x_i ) for all ( i ).Conclusion:[boxed{f(x) = x text{ for all } x in {x_1, x_2, cdots, x_n}}]This is the unique solution satisfying the given condition.

🧠:Okay, let's see. So the problem is about Cindy leaving school and cycling home at different speeds, resulting in different arrival times. We need to find the speed she needs to cycle to arrive home at 5:00 PM. The answer choices are given, so maybe we can use some formula involving distance, speed, and time. First, let me recall that distance equals speed multiplied by time, right? So, D = S * T. Maybe we can use that here. The problem says that if she cycles at 20 km/h, she arrives home at 4:30 PM, and if she cycles at 10 km/h, she arrives home at 5:15 PM. We need to find the speed required to arrive at 5:00 PM. Hmm. So the distance from school to home should be the same in both cases, right? Because it's the same route. So the distance when she cycles at 20 km/h is the same as when she cycles at 10 km/h. The difference is the time taken, which affects her arrival time. Let me denote the time she takes to cycle home when she goes at 20 km/h as t hours. Then, when she cycles at 10 km/h, it would take her longer. Specifically, the time difference between arriving at 4:30 PM and 5:15 PM is 45 minutes, which is 0.75 hours. Wait, 45 minutes is 0.75 hours? Let me check. 60 minutes is 1 hour, so 45 minutes is 45/60 = 0.75 hours. Right. But wait, actually, when she cycles slower, she takes more time. So if she arrives 45 minutes later when going at 10 km/h compared to 20 km/h, the time difference is 45 minutes. So the time taken at 10 km/h is t + 0.75 hours. But we also know that the distance is the same, so we can set up two equations. At 20 km/h: D = 20 * tAt 10 km/h: D = 10 * (t + 0.75)Since both equal D, we can set them equal to each other:20t = 10(t + 0.75)Let me solve this equation. 20t = 10t + 7.5Subtract 10t from both sides:10t = 7.5Divide both sides by 10:t = 0.75 hoursWait, 0.75 hours is 45 minutes. Hmm. So that means when she cycles at 20 km/h, she takes 45 minutes to get home, arriving at 4:30 PM. So if her cycling time is 45 minutes at 20 km/h, then she leaves school at 4:30 PM minus 45 minutes, which is 3:45 PM. Wait, but does the problem say she leaves school at the same time every day? Yes. So regardless of her speed, she leaves school at the same time. Therefore, the departure time is fixed. So when she cycles faster, she arrives earlier, and slower, arrives later. So if she leaves school at 3:45 PM, cycling at 20 km/h takes 45 minutes (0.75 hours) to get home at 4:30 PM. If she cycles at 10 km/h, the time taken would be t + 0.75 hours? Wait, no. Wait, the first time we found t was 0.75 hours. But when she cycles at 10 km/h, the time is t + 0.75? Wait, maybe I confused the variables. Let me re-examine. Wait, perhaps the time difference between arriving at 4:30 and 5:15 is 45 minutes, so the slower trip takes 45 minutes longer. So if the faster trip takes t hours, the slower trip takes t + 0.75 hours. Then, since the distance is the same:20t = 10(t + 0.75)So solving:20t = 10t + 7.510t = 7.5t = 0.75 hours, which is 45 minutes. So that checks out. So when she cycles at 20 km/h, it takes 45 minutes, arriving at 4:30 PM. Therefore, she leaves school at 4:30 PM minus 45 minutes, which is 3:45 PM. Similarly, when she cycles at 10 km/h, it takes t + 0.75 = 0.75 + 0.75 = 1.5 hours, which is 90 minutes, arriving at 5:15 PM. Let me check the departure time: 5:15 PM minus 1.5 hours (90 minutes) is 3:45 PM. So that matches. So the distance is 20 km/h * 0.75 h = 15 km. Let's check: 10 km/h * 1.5 h = 15 km. Correct. So the distance from school to home is 15 km. Now, the question is, at what speed must she cycle to arrive home at 5:00 PM? So her departure time is 3:45 PM, and she wants to arrive at 5:00 PM. So the total time she wants to take is from 3:45 PM to 5:00 PM. Let's compute that time. From 3:45 PM to 4:45 PM is 1 hour, and from 4:45 PM to 5:00 PM is 15 minutes. So total time is 1 hour 15 minutes, which is 1.25 hours. So she needs to cycle 15 km in 1.25 hours. Therefore, her required speed is distance divided by time: 15 km / 1.25 h. Calculating that: 15 / 1.25. Let's see. 1.25 is 5/4, so dividing by 5/4 is multiplying by 4/5. So 15 * 4/5 = 12. So 12 km/h? Wait, but that's answer choice (D). But wait, let me check again. Wait, if the distance is 15 km and the time needed is 1.25 hours, then speed is 15 / 1.25. Let's compute that. 15 divided by 1.25. 1.25 goes into 15 how many times? 1.25 * 10 = 12.5, which is less than 15. 1.25 * 12 = 15. Exactly. So 15 / 1.25 = 12. So 12 km/h. So answer choice (D). But wait, the answer choices include 16 2/3, 15, 13 1/3, 12, and 18 3/4. So 12 is (D). But let me check once more to be sure. Wait, perhaps I made a mistake in calculating the required time. Let's verify the time she needs. She leaves school at 3:45 PM and wants to arrive at 5:00 PM. So from 3:45 to 5:00 is 1 hour and 15 minutes, which is 1.25 hours. Yes. Distance is 15 km. So speed is 15 / 1.25. 15 divided by 1.25. Let's compute:1.25 * 10 = 12.515 - 12.5 = 2.5So 10 gives 12.5, and we need 15. So remaining 2.5. 1.25 * 2 = 2.5. So total time is 10 + 2 = 12 minutes? Wait, no. Wait, no, that's if we think in terms of units. Wait, maybe better to do decimal division. Alternatively, 15 / 1.25: multiply numerator and denominator by 100 to eliminate decimals: 1500 / 125. 125 * 12 = 1500. So 1500 / 125 = 12. So 12 km/h. So answer is D. But wait, let me think again. The answer seems straightforward, but let me confirm if the logic is correct. We established the distance is 15 km. To arrive at 5:00 PM, which is 1 hour 15 minutes after 3:45 PM, so total time is 1.25 hours. Thus, speed is 15 / 1.25 = 12. So yes, 12 km/h. But wait, in the answer choices, there is an option (C) 13 1/3. Maybe I did something wrong. Let me check again. Wait, perhaps I made a mistake in computing the time difference. Let me recheck the departure time. When she cycles at 20 km/h, arrives at 4:30 PM. The time taken is 45 minutes (0.75 hours). Therefore, departure time is 4:30 PM minus 45 minutes, which is 3:45 PM. Correct. When she cycles at 10 km/h, arrives at 5:15 PM. Time taken is 1.5 hours (15 km /10 km/h). 1.5 hours is 90 minutes. 5:15 PM minus 90 minutes is 3:45 PM. Correct. Therefore, departure time is 3:45 PM. So to arrive at 5:00 PM, the time taken is 1 hour 15 minutes, which is 1.25 hours. Therefore, speed needed is 15 km / 1.25 hours = 12 km/h. So answer is D. But maybe I should check if the answer is not supposed to be 13 1/3. Wait, let me think differently. Suppose someone might approach this problem using the time difference. The time difference between 4:30 PM and 5:15 PM is 45 minutes, which is 0.75 hours. Let’s denote the distance as D. At 20 km/h, time taken is D / 20. At 10 km/h, time taken is D / 10. The difference between these times is 45 minutes = 0.75 hours. Therefore, D /10 - D /20 = 0.75Compute this: (2D - D)/20 = 0.75D /20 = 0.75Multiply both sides by 20: D = 15 km. So that's the same as before. So distance is 15 km. Then, to find the speed needed to arrive at 5:00 PM, which is halfway between 4:30 and 5:15? Wait, no. The desired arrival time is 5:00 PM. The time taken when arriving at 5:00 PM is T = 5:00 PM - 3:45 PM = 1 hour 15 minutes = 1.25 hours. So speed S = 15 km /1.25 hours = 12 km/h. So answer is D. Therefore, 12 km/h is correct. But why is option C 13 1/3? Maybe a common mistake would be miscalculating the time difference. Let me check if someone could think the time difference is different. Alternatively, maybe the problem is being approached by considering the difference in arrival times and relating it to speed. Alternatively, think in terms of harmonic mean? Because we have two different speeds and want an average? But harmonic mean of 10 and 20 is 2*10*20/(10+20) = 400/30 ≈ 13.33, which is 13 1/3. But that's answer C. Wait, but why harmonic mean? That's for average speed over the same distance. But in this problem, we are not looking for an average speed. We have a fixed distance and want the speed that gets her home at a specific time. So harmonic mean is not applicable here. Alternatively, maybe someone could set up the problem incorrectly. Suppose they thought the time difference is 30 minutes instead of 45. Let me see. Wait, arrival times are 4:30 and 5:15. The difference is 45 minutes. So that's correct. Alternatively, perhaps someone miscalculates the departure time. Suppose they thought departure time is 4:30 minus t, where t is the time taken at 20 km/h, but miscalculated t. But in our calculation, we found t=0.75 hours, which is correct. Alternatively, maybe someone thought the total time from school to home is different. But our calculation of the distance as 15 km seems correct. Wait, but let me check with another approach. Let’s denote the departure time as T, and the arrival times as T + t1 = 4:30 PM and T + t2 =5:15 PM. Given that t2 - t1 = 45 minutes. Since D = 20*t1 =10*t2. Therefore, 20*t1 =10*t2 => 2*t1 = t2. And t2 - t1 = 45 minutes = 0.75 hours. So substituting t2 = 2*t1 into t2 - t1 = 0.75:2*t1 - t1 = t1 = 0.75 hours. Therefore, t1 = 0.75 hours, t2 = 1.5 hours. Thus, departure time T = 4:30 PM - 0.75 hours = 3:45 PM. Then, desired arrival time is 5:00 PM, so time taken is 5:00 PM - 3:45 PM = 1.25 hours. Thus, speed S = D /1.25 =15 /1.25 =12. Same result. Therefore, answer is D)12. So why is option C present? Maybe it's a trap if someone does the harmonic mean. Or, perhaps, miscalculates 15 divided by 1.25. Let me check 15 divided by 1.25 again. 1.25 multiplied by 10 is 12.5. 1.25*12=15. So 15/1.25=12. Correct. Alternatively, maybe someone thought the time difference is 1 hour (from 4:30 to 5:30?), but the problem states 5:15, so no. Alternatively, maybe they thought the distance is 20 km/h *0.75 h=15 km, then to find the time to arrive at 5:00, which is 15 minutes later than 4:45? Wait, no. Let's not confuse. Wait, perhaps someone could have thought that the desired arrival time is halfway between 4:30 and 5:15, but 5:00 is 30 minutes after 4:30 and 15 minutes before 5:15, so it's not halfway. The time between 4:30 and 5:15 is 45 minutes, so halfway would be 22.5 minutes, which is 4:52:30 PM. But we need 5:00 PM. Alternatively, maybe using the formula for two speeds and two times, find the required speed. But I think the way we approached is correct. Found the distance, found the required time, calculated speed. So answer is 12 km/h, option D. Therefore, the correct answer should be D)12.Final Answerboxed{D}

✅:Cindy leaves school at the same time every day. If she cycles at (20 text{km/h}), she arrives home at 4:30 PM. If she cycles at (10 text{km/h}), she arrives home at 5:15 PM. We need to find the speed required for her to arrive home at 5:00 PM.1. Determine the distance to home: - Let (t) be the time taken to cycle home at 20 km/h. - The time taken to cycle home at 10 km/h is (t + 0.75) hours (since the difference in arrival times is 45 minutes, which is 0.75 hours). - The distance (D) is the same for both speeds: [ 20t = 10(t + 0.75) ] - Solving for (t): [ 20t = 10t + 7.5 implies 10t = 7.5 implies t = 0.75 text{hours} ] - Therefore, the distance (D) is: [ D = 20 times 0.75 = 15 text{km} ]2. Determine the required speed to arrive at 5:00 PM: - Cindy leaves school at 3:45 PM (since 4:30 PM minus 45 minutes is 3:45 PM). - The time needed to arrive at 5:00 PM is 1 hour and 15 minutes, which is 1.25 hours. - The required speed (S) is: [ S = frac{15 text{km}}{1.25 text{hours}} = 12 text{km/h} ]Thus, the speed Cindy must cycle to arrive home at 5:00 PM is (boxed{D}).

🔑:1. Introduce Variables and Distance Calculation: Let ( d ) represent the distance from Cindy's school to home in kilometers.2. Cycling at 20 km/h: - Let ( t ) be the time in hours that Cindy takes to travel from school to home at ( 20 text{km/h} ). - The time equation when cycling at ( 20 text{km/h} ) is: [ d = 20t ]3. Cycling at 10 km/h: - Cindy takes ( frac{3}{4} ) hour longer when cycling at ( 10 text{km/h} ), so the time taken is ( t + frac{3}{4} ) hours. - The distance equation when cycling at ( 10 text{km/h} ) is: [ d = 10 left(t + frac{3}{4}right) ]4. Equating Both Distances: Since the distance ( d ) is the same in both cases: [ 20t = 10 left(t + frac{3}{4}right) ]5. Solve for ( t ): - Expand and simplify the equation: [ 20t = 10t + frac{30}{4} ] [ 20t = 10t + 7.5 ] [ 10t = 7.5 ] [ t = frac{7.5}{10} = frac{3}{4} text{hours} ]6. Calculate Distance ( d ): - Using ( d = 20t ): [ d = 20 times frac{3}{4} = 15 text{km} ]7. Cycling to Arrive at 5:00 p.m.: - If Cindy needs to arrive home at 5:00 p.m., she travels for: [ left(frac{3}{4} + frac{1}{2}right) text{hours} ] [ = left(frac{3}{4} + frac{2}{4}right) text{hours} = frac{5}{4} text{hours} ]8. Calculate the Required Speed: - The speed ( s ) needed to travel 15 km in ( frac{5}{4} ) hours: [ s = frac{d}{t} = frac{15}{frac{5}{4}} ] [ s = 15 times frac{4}{5} = 12 text{km/h} ] Conclusion:[boxed{12 text{km/h}}]

🧠:Okay, so I need to find the area of quadrilateral ABCD given that the area of triangle ABC is 2019. The problem mentions that M and N are midpoints of BC and AD, respectively. Then, the segments MN and AC intersect at point O such that MO = ON. Hmm, let me visualize this first. Maybe drawing a diagram would help.Alright, let me sketch quadrilateral ABCD. Let's label the vertices in order: A, B, C, D. M is the midpoint of BC, so BM = MC. N is the midpoint of AD, so AN = ND. Then, connecting M and N with segment MN. This segment intersects the diagonal AC at point O, and it's given that MO = ON. So O is the midpoint of MN. Interesting.Given that the area of triangle ABC is 2019. We need to find the area of the entire quadrilateral ABCD. Quadrilaterals can be tricky because their area depends on the specific shape, but maybe there's a way to relate the given information to the areas of other parts.Since M and N are midpoints, perhaps there's a midline theorem application here. The midline theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. But here, we have a quadrilateral. Maybe if I split the quadrilateral into triangles and analyze their areas?First, let's note that the area of triangle ABC is given. If we can find the area of triangle ADC, then the total area of the quadrilateral would be the sum of the areas of ABC and ADC. So, maybe I need to find the area of ADC in terms of the given area.But how does the midpoint information help here? MN connects midpoints of BC and AD. And O is the midpoint of MN, which lies on AC. So AC is divided by O into two parts. Since O is the midpoint of MN, and MN connects the midpoints of BC and AD, perhaps O has a special relationship with AC.Let me consider coordinate geometry. Maybe assigning coordinates to the points would help. Let me try that.Let's place point A at (0, 0) to simplify. Let’s assign coordinates to the other points. Let’s denote:- A = (0, 0)- B = (2b, 0) [placing B on the x-axis for simplicity]- D = (2d, 2e) [since N is the midpoint of AD, so N would be at (d, e)]- C is another point; since M is the midpoint of BC, let's denote C = (2c, 2f), so M would be the midpoint of B(2b, 0) and C(2c, 2f), so M = (b + c, f)Wait, maybe coordinates can get messy, but perhaps manageable.But maybe there's a better approach. Let's think about vectors or area ratios.Since O is the midpoint of MN, and MN connects midpoints M and N. So MN is the line connecting M (midpoint of BC) and N (midpoint of AD). The intersection point O with AC divides AC in some ratio. Since MO = ON, O is the midpoint of MN. So maybe AC is divided by O in a particular ratio.Alternatively, perhaps using the theorem of intersecting lines in a quadrilateral. Wait, maybe using the concept of centroids or harmonic division?Alternatively, think of ABCD as two triangles, ABC and ADC, as mentioned before. If I can relate the area of ADC to ABC, that would solve the problem. But how?Given that M is the midpoint of BC and N is the midpoint of AD, MN connects these midpoints. If O is the midpoint of MN, which is on AC, maybe the position of O on AC can be determined. If we can find the ratio of AO to OC, then perhaps we can relate the areas.Let me denote the coordinates more carefully.Let’s set coordinate system with point A at (0,0), D at (2d, 2e), so midpoint N is at (d, e). Let’s set point B at (2b, 0), so midpoint of BC is M. Let’s let point C be (2c, 2f). Therefore, midpoint M of BC is [(2b + 2c)/2, (0 + 2f)/2] = (b + c, f).Then, segment MN connects N(d, e) and M(b + c, f). The midpoint O of MN is [(d + b + c)/2, (e + f)/2]. This point O lies on AC. Let's parametrize AC. Since A is (0,0) and C is (2c, 2f), the parametric equation of AC is (2c*t, 2f*t) for t in [0,1]. So point O must satisfy:(2c*t, 2f*t) = [(d + b + c)/2, (e + f)/2]Therefore, equating coordinates:2c*t = (d + b + c)/2 => t = (d + b + c)/(4c)Similarly, 2f*t = (e + f)/2 => t = (e + f)/(4f)Therefore, these two expressions for t must be equal:(d + b + c)/(4c) = (e + f)/(4f)Simplify:(d + b + c)/c = (e + f)/fMultiply both sides by 4f:(d + b + c)f = (e + f)cSo, expanding:df + bf + cf = ec + fcSubtract fc from both sides:df + bf = ecSo, we have:df + bf = ecHmm, this is an equation relating the coordinates. Let me note this.But how does this help us find the area?The area of triangle ABC is 2019. In coordinates, the area can be computed as:Area ABC = 1/2 | (B × C) | since A is at origin.Wait, if A is (0,0), B is (2b, 0), and C is (2c, 2f), then the area is 1/2 | (2b*(2f) - 0*(2c - 0)) | = 1/2 |4bf| = 2bf.Given that the area is 2019, so 2bf = 2019 => bf = 2019/2.Hmm, so bf = 1009.5.Now, the area of quadrilateral ABCD is the sum of areas of ABC and ADC. Area ABC is 2019, so we need area ADC.Area ADC can be computed as 1/2 |A × D + D × C + C × A|, but since A is (0,0), it simplifies to 1/2 | (0*(2e) - 0*(2d)) + (2d*(2f) - 2e*(2c)) + (2c*0 - 2f*0) | = 1/2 |0 + (4df - 4ec) + 0| = 1/2 |4df - 4ec| = 2 |df - ec|.From earlier, we had df + bf = ec. So, df - ec = -bf.Therefore, area ADC = 2 |df - ec| = 2 | -bf | = 2 |bf|.But bf is 1009.5, so area ADC = 2 * 1009.5 = 2019.Therefore, total area ABCD = 2019 + 2019 = 4038.Wait, so the area of ADC is also 2019? So the quadrilateral has area 4038. Is that possible?But let me check again. From the coordinate setup:Area ABC = 2bf = 2019, so bf = 1009.5.Then, from the equation df + bf = ec, so df - ec = -bf.Therefore, |df - ec| = | -bf | = bf.Thus, area ADC = 2 * bf = 2 * 1009.5 = 2019.Therefore, total area ABCD = 2019 + 2019 = 4038.Hmm, that seems straightforward. But let me confirm if this is consistent.Alternatively, maybe there's a symmetry here. Since O is the midpoint of MN and lies on AC, maybe the diagonal AC is being divided by O in such a way that the areas of ABC and ADC are equal. But if O is the midpoint of MN, does that imply something about the areas?Alternatively, using vectors. Let's see.Let’s denote vectors with position vectors from A as the origin.Let’s let vector AB = b, AD = d. Then, point B is at vector b, D is at vector d. Then, midpoint N is at (d)/2.Point C can be expressed as some vector. Wait, maybe this approach is more complicated. Let me stick with the coordinate system.Alternatively, since M is the midpoint of BC and N is the midpoint of AD, then MN connects midpoints. In a quadrilateral, the line connecting midpoints of two sides is related to the midline, but in general, the midline of a quadrilateral is the segment connecting the midpoints of two opposite sides and is parallel to the other two sides and equal in length to half their sum. However, here MN connects midpoints of BC and AD, which are adjacent sides. Hmm, not the usual midline.But perhaps in this problem, the key is that O is the midpoint of MN and lies on AC. From the coordinate analysis, this leads to a relationship that allows the area of ADC to be equal to the area of ABC. Therefore, the total area is double the area of ABC, so 4038.But let me check with another approach to confirm.Suppose we use mass point geometry. If O is the midpoint of MN, then considering the masses at M and N would each be 1, so O is their center. Since O lies on AC, maybe we can find the ratios along AC.Alternatively, using coordinate geometry again, but maybe assigning specific coordinates for simplicity.Let me take a specific case where ABCD is a parallelogram. Wait, but in a parallelogram, the midpoints M and N would be such that MN connects the midpoints of BC and AD. But in a parallelogram, BC and AD are congruent and parallel. Then, the midpoint of BC and midpoint of AD would coincide if it's a parallelogram, but no. Wait, in a parallelogram, AD is congruent and parallel to BC. Wait, no, in a parallelogram, AB is congruent and parallel to CD, and AD congruent and parallel to BC. So, if ABCD is a parallelogram, then M is the midpoint of BC, N is the midpoint of AD. Then MN would connect midpoints of two sides. In a parallelogram, the line connecting midpoints of two sides would be parallel to the other sides and half their length? Not sure. Maybe in this case, MN would pass through the center of the parallelogram. Then O would be the center, which is also the midpoint of AC. Therefore, in a parallelogram, O would be the midpoint of AC as well. Then MO = ON, which is true since O is the midpoint. But in a parallelogram, the area of ABC is half the area of the parallelogram. So if the area of ABC is 2019, the total area would be 4038. Which matches the previous result. So in the case of a parallelogram, the area is double that of ABC. But the problem states a general convex quadrilateral. So even in a general case, the area is 4038?Wait, so in the coordinate system, regardless of the specific coordinates (as long as they satisfy the given midpoint conditions and O being the midpoint on AC), the area comes out as 4038. So maybe the answer is 4038.But why is that? How come the area of ADC equals the area of ABC in this configuration? Is there a more intuitive explanation?Perhaps using the concept of midpoints and the midline. Let's think of triangle ABC. Since M is the midpoint of BC, and N is the midpoint of AD, and O is the midpoint of MN. Then, connecting these midpoints, perhaps there's a relation that averages the areas.Alternatively, consider that the line MN is connecting midpoints, and O is the midpoint of MN on AC. So maybe the line MN divides the quadrilateral into regions whose areas are related. But not sure.Alternatively, using affine transformations. Since affine transformations preserve ratios and midpoints, we can transform the quadrilateral into a simpler form, like a parallelogram, without changing the area ratios. Then, as in the parallelogram case, the area would be double that of ABC. Therefore, 4038.Alternatively, another approach: Since M and N are midpoints, then MN is a line connecting them. O is the midpoint of MN, so it's like the center of mass if equal weights are at M and N.Since O is on AC, this imposes a condition on the position of AC relative to M and N.But in the coordinate system, we found that the area of ADC equals the area of ABC. Hence, the total area is 4038.Given that multiple approaches lead to the same result, I feel confident that the answer is 4038.Final Answerboxed{4038}

✅:Given a convex quadrilateral (ABCD) with midpoints (M) on (BC) and (N) on (AD), such that the segments (MN) and (AC) intersect at point (O) where (MO = ON). The area of triangle (ABC) is known to be 2019. We need to find the area of quadrilateral (ABCD).1. Coordinate Setup: - Assign coordinates: (A(0, 0)), (B(2b, 0)), (D(2d, 2e)), and (C(2c, 2f)). - Midpoints: (M(b+c, f)) (midpoint of (BC)) and (N(d, e)) (midpoint of (AD)).2. Midpoint (O) on (AC): - (O) is the midpoint of (MN), so coordinates of (O) are (left(frac{d+b+c}{2}, frac{e+f}{2}right)). - (O) lies on (AC), parameterized as ((2ct, 2ft)) for (t in [0, 1]).3. Equating Coordinates: - Solve for (t) in terms of coordinates: [ 2ct = frac{d + b + c}{2} quad text{and} quad 2ft = frac{e + f}{2} ] - Equate the expressions for (t): [ frac{d + b + c}{4c} = frac{e + f}{4f} ] - Simplify to get (df + bf = ec).4. Area Calculations: - Area of (triangle ABC) is (2bf = 2019), so (bf = 1009.5). - Area of (triangle ADC) is calculated using the determinant method: [ text{Area of } triangle ADC = 2|df - ec| = 2| -bf | = 2 times 1009.5 = 2019 ]5. Total Area: - Total area of quadrilateral (ABCD) is the sum of areas of (triangle ABC) and (triangle ADC): [ 2019 + 2019 = 4038 ]Thus, the area of the quadrilateral (ABCD) is (boxed{4038}).

🔑:1. Identify the midpoints: Let ( K ) and ( L ) be the midpoints of the sides ( AB ) and ( CD ) of the convex quadrilateral ( ABCD ) respectively. 2. Form the parallelogram: Since ( K ), ( M ), ( L ), and ( N ) are the midpoints of the sides of quadrilateral ( ABCD ), the quadrilateral ( KMLN ) is a parallelogram by the midsegment theorem (Theorem: A quadrilateral formed by joining the midpoints of the sides of another quadrilateral is a parallelogram).3. Diagonal intersection: The diagonals ( KM ) and ( LN ) of the parallelogram ( KMLN ) intersect at point ( O ), and we are given that ( MO = ON ).4. Equal areas in triangles: We know from the properties of parallelograms that the diagonals bisect each other into equal halves, hence ( KO = OL ) and ( MO = ON ). Also, since ( O ) is the midpoint of the diagonals, the areas of triangles ( OKM ) and ( ONL ) are equal. Let's denote the area of triangle ( BKM ) as ( S ). By symmetry and equality of heights and bases, [ S_{OKM} = S_{NOL} = S ]. Therefore, the areas of triangles ( DNL ) and ( ONL ) are also equal: [ S_{DNL} = S_{ONL} = S ].5. Relating areas via similarity: Since ( BK ) and ( KM ) are midsegments of ( triangle ABC ), the triangles ( BKM ) and ( ABC ) are similar with a ratio of (frac{1}{2}). Therefore, [ S_{ABC} = 4S_{BKM} = 4S ]. Similarly, [ S_{ADC} = 4S_{DNL} = 4S ].6. Adding areas: Since ( S_{ABC} = 2019 ) and this ( S_{ADC} ) also equals the same value due to the symmetry and same area property, we hence calculate: [ S_{ABCD} = S_{ABC} + S_{ADC} = 2019 + 2019 = 4038 ].7. Conclusion: Hence, the area of the quadrilateral ( ABCD ) is ( boxed{4038} ).

🧠:Okay, so I have this equation to solve: [x - 1/2] = 3x - 5. The brackets here mean the greatest integer less than or equal to x - 1/2. Alright, so I need to find the value of x that satisfies this equation. Let me think step by step how to approach this.First, I remember that the floor function, denoted by [x], gives the greatest integer less than or equal to x. So, in this case, [x - 1/2] is the floor of (x - 0.5). The right side of the equation is 3x - 5. So, I need to find x such that when you take the floor of (x - 0.5), it equals 3x - 5.Hmm, let me recall that for any real number y, [y] ≤ y < [y] + 1. That is, the floor of y is the integer part of y, and the next integer is one more than that. So, maybe I can use this property here. Let's let y = x - 1/2. Then, [y] = 3x - 5. So, according to the floor function properties:[y] ≤ y < [y] + 1Substituting back in terms of x:3x - 5 ≤ x - 1/2 < 3x - 5 + 1Simplifying the inequalities:Left inequality: 3x - 5 ≤ x - 1/2Right inequality: x - 1/2 < 3x - 5 + 1 => x - 1/2 < 3x - 4Let me solve the left inequality first:3x - 5 ≤ x - 1/2Subtract x from both sides:2x - 5 ≤ -1/2Add 5 to both sides:2x ≤ 9/2Divide by 2:x ≤ 9/4, which is 2.25Now the right inequality:x - 1/2 < 3x - 4Subtract x from both sides:-1/2 < 2x - 4Add 4 to both sides:7/2 < 2xDivide by 2:7/4 < x, which is 1.75So combining both inequalities, we have:7/4 < x ≤ 9/4So x is in the interval (1.75, 2.25]But wait, the floor function [x - 1/2] is equal to 3x -5, which must be an integer since the floor function always yields an integer. Therefore, 3x -5 must be an integer. Let's denote that integer as k. So, 3x -5 = k, where k is an integer. Then, x = (k +5)/3.So, x must be equal to (k +5)/3 for some integer k. Additionally, x must be in the interval (1.75, 2.25]. Let's substitute x = (k +5)/3 into the interval:1.75 < (k +5)/3 ≤ 2.25Multiply all parts by 3:5.25 < k +5 ≤ 6.75Subtract 5 from all parts:0.25 < k ≤ 1.75But k must be an integer. The integers between 0.25 and 1.75 are k = 1. So, k =1 is the only integer in that interval. Therefore, x = (1 +5)/3 =6/3=2.Wait, so x=2? Let's check if this solution works in the original equation.Left side: [x - 1/2] = [2 - 0.5] = [1.5] =1Right side: 3x -5=3*2 -5=6-5=1Yes, both sides equal 1. So x=2 is a solution.But let me check if there are other possible k values. Wait, in the interval for k, 0.25 <k ≤1.75, k must be integer. The only integer is 1. So, only k=1. So x=2 is the only solution. But let me check if x=2 is within the original interval (1.75,2.25]. Yes, 2 is in that interval.But let me also check if maybe there's another solution if I consider different k's. Wait, suppose k=0. Then x=(0+5)/3≈1.666..., which is less than 1.75, so not in the interval. If k=2, then x=(2+5)/3≈2.333..., which is more than 2.25, so also outside the interval. So only k=1 gives x=2. So x=2 is the only solution.But let me confirm again. Let's suppose there is another x in (1.75,2.25] that might satisfy [x - 1/2] =3x -5. Let's test x=2.25. Then left side: [2.25 -0.5]=[1.75]=1. Right side:3*2.25 -5=6.75 -5=1.75. But 1 ≠1.75, so x=2.25 is not a solution.How about x=1.8? Then left side: [1.8 -0.5]=[1.3]=1. Right side:3*1.8 -5=5.4 -5=0.4. But 1≠0.4, so not equal.Wait, maybe x=2 is the only solution? Let's check another point. Let's take x=2. Then as before, both sides equal 1. If x=2.2, left side: [2.2 -0.5]=[1.7]=1. Right side:3*2.2 -5=6.6 -5=1.6. Not equal. If x=1.9, left side: [1.4]=1, right side: 3*1.9 -5=5.7-5=0.7. Not equal.How about x=2.0? Which we already checked. So, seems like x=2 is the only solution. Wait, but let me check the original inequality again. We found x must be in (1.75,2.25], and x must be (k+5)/3 where k is integer. So k=1 gives x=2, which is in that interval. Any other k? k=1 is the only integer in 0.25<k<=1.75. So no other k's. Therefore, x=2 is the only solution.But let me go back to the original equation and see if there's another way to approach this. Let's denote that [x - 1/2] =3x -5. Let me call the left side as an integer, say n. So n =3x -5. Then x=(n +5)/3. Also, since n = [x -1/2], then n ≤x -1/2 <n +1.Substituting x=(n +5)/3 into the inequality:n ≤ (n +5)/3 -1/2 <n +1Let me compute the middle term:(n +5)/3 -1/2 = (2(n +5) -3)/6 = (2n +10 -3)/6 = (2n +7)/6So the inequality becomes:n ≤ (2n +7)/6 <n +1Multiply all parts by 6:6n ≤ 2n +7 <6n +6Left inequality:6n ≤2n +7 → 4n ≤7 →n ≤7/4=1.75Right inequality:2n +7 <6n +6 →7 -6 <6n -2n →1 <4n →n >1/4=0.25So combining both inequalities:0.25 <n ≤1.75But n must be an integer. So n can be 1, since 0.25 <1 ≤1.75. So n=1. Therefore, x=(1 +5)/3=6/3=2.Therefore, x=2 is the solution.So, seems like the only solution is x=2. Let me check once again:Left side: [2 -1/2] = [1.5] =1Right side:3*2 -5=6 -5=1Yes, both sides equal 1. Therefore, x=2 is indeed the solution.But let me just ensure that there are no other possible values. For example, suppose someone thinks that maybe x=1.5. Let's check:Left side: [1.5 -0.5] =[1]=1Right side:3*1.5 -5=4.5 -5= -0.51≠-0.5, so that's not a solution. How about x=2.5? Left side: [2.5 -0.5]=[2]=2. Right side:3*2.5 -5=7.5 -5=2.5. 2≠2.5. Not equal. How about x=1. Let's check:Left side: [1 -0.5]=[0.5]=0Right side:3*1 -5= -2. 0≠-2. Nope.So, indeed, x=2 is the only solution.I think that's thorough. So the answer is x=2.

✅:The given equation is (leftlfloor x - frac{1}{2} rightrfloor = 3x - 5), where (lfloor cdot rfloor) denotes the floor function. To solve this:1. Define (k = 3x - 5): Since the floor function yields an integer, (3x - 5) must be an integer (k). Thus, (x = frac{k + 5}{3}).2. Apply floor function properties: For (leftlfloor x - frac{1}{2} rightrfloor = k), the inequalities (k leq x - frac{1}{2} < k + 1) must hold. Substituting (x = frac{k + 5}{3}): [ k leq frac{k + 5}{3} - frac{1}{2} < k + 1 ] Simplifying the middle term: [ frac{k + 5}{3} - frac{1}{2} = frac{2k + 10 - 3}{6} = frac{2k + 7}{6} ] This leads to: [ k leq frac{2k + 7}{6} < k + 1 ] Multiplying through by 6: [ 6k leq 2k + 7 < 6k + 6 ] Solving the inequalities: - Left: (4k leq 7 implies k leq frac{7}{4} = 1.75) - Right: (7 < 4k + 6 implies 1 < 4k implies k > frac{1}{4} = 0.25)3. Determine valid integer (k): The only integer (k) in (0.25 < k leq 1.75) is (k = 1).4. Solve for (x): Substituting (k = 1) into (x = frac{k + 5}{3}): [ x = frac{1 + 5}{3} = 2 ]5. Verification: Check (x = 2) in the original equation: - Left side: (leftlfloor 2 - frac{1}{2} rightrfloor = leftlfloor 1.5 rightrfloor = 1) - Right side: (3(2) - 5 = 1)Both sides equal 1, confirming the solution.Final Answer: (boxed{2})

🔑:1. Understanding the Floor Function: The floor function ( [x] ) represents the greatest integer less than or equal to ( x ). Therefore, ( left[x - frac{1}{2}right] ) represents the greatest integer less than or equal to ( x - frac{1}{2} ).2. Setting Up the Equation: We need to solve the equation ( left[x - frac{1}{2}right] = 3x - 5 ).3. Inequality Setup: For the equation to hold, we must have: [ left[x - frac{1}{2}right] leq x - frac{1}{2} ] This is always true because the floor function, by definition, is less than or equal to the number itself.4. Integer Check: Since ( 3x - 5 ) is an integer (because it equals the floor function), we need: [ 3x - 5 leq x - frac{1}{2} ] Rearranging the inequality, we get: [ 3x - 5 leq x - frac{1}{2} implies 2x leq frac{9}{2} implies x leq frac{9}{4} ] Since ( x leq frac{9}{4} ), we need to check values of ( x ) that satisfy this inequality and also make ( 3x - 5 ) an integer.5. Check for Integer Solutions: - Let's consider ( x = 2 ) (Integer value close to (frac{9}{4})): [ 3 cdot 2 - 5 = 6 - 5 = 1 ] Since ( left[2 - frac{1}{2}right] = left[frac{3}{2}right] = 1 ), the values match, confirming that ( x = 2 ) satisfies the equation.6. Verification: - Verify the condition for another integer less than ( frac{9}{4} ): - If ( x = 3 ): [ 3 cdot 3 - 5 = 9 - 5 = 4 ] However, ( left[3 - frac{1}{2}right] = left[frac{5}{2}right] = 2 ), which does not match. Hence, ( x = 2 ) is the only solution. Conclusion:[boxed{2}]