🧠:Okay, let's try to figure out which integer is closest to 99 multiplied by 9. The options are A) 10,000, B) 100, C) 100,000, D) 1,000, and E) 10. Hmm, first, I need to compute 99 times 9. Let me think about how to do this. Maybe I can do it step by step.Wait, 99 times 9. Let me break it down. 100 times 9 is 900, right? Because 100*9 is straightforward. But since it's 99, that's one less than 100. So, 100*9 is 900, subtract 1*9 which is 9. So, 900 - 9 = 891. So, 99*9 equals 891. Now, I need to find which of the given options is closest to 891.Looking at the options: A is 10,000, which is way bigger. B is 100, which is 100. C is 100,000, even bigger. D is 1,000, and E is 10. So, comparing 891 to these options. Let's see how far each is from 891.First, let's compute the differences:For option A) 10,000: 10,000 - 891 = 9,109. That's a huge difference.Option B) 100: 891 - 100 = 791. Still a big difference, but smaller than A.Option C) 100,000: 100,000 - 891 = 99,109. Even larger. Not close at all.Option D) 1,000: 1,000 - 891 = 109. So, the difference here is 109.Option E) 10: 891 - 10 = 881. That's the largest difference yet.So, the differences are:A) 9,109B) 791C) 99,109D) 109E) 881The smallest difference is option D with 109. So, 1,000 is 109 away from 891, which is the closest among all the options. Therefore, the answer should be D) 1000.Wait, but let me double-check my calculation of 99*9 to make sure I didn't make a mistake. 99*9. Let me compute it another way. 9 times 100 is 900, subtract 9 gives 891. Yes, that's correct. Alternatively, breaking 99 into 90 + 9. So, 90*9 = 810, and 9*9=81. Adding 810 +81 = 891. Yep, same result. So 891 is correct.Then, the closest option is 1,000, which is 109 away. The next closest is 100, which is 791 away. So, definitely D is the closest. So the answer is D.

✅:The integer closest to ( 99 times 9 ) can be determined by calculating the product and comparing it to the given options. 1. Calculate ( 99 times 9 ): - Recognize that ( 99 = 100 - 1 ). - Compute ( 100 times 9 = 900 ), then subtract ( 1 times 9 = 9 ). - Result: ( 900 - 9 = 891 ). 2. Compare 891 to the options: - Compute absolute differences: - ( |10,!000 - 891| = 9,!109 ) - ( |100 - 891| = 791 ) - ( |100,!000 - 891| = 99,!109 ) - ( |1,!000 - 891| = 109 ) - ( |10 - 891| = 881 ). - The smallest difference is 109, corresponding to 1,000. Final Answer: (boxed{D})

🔑:1. First, let's evaluate the product (99 times 9): [ 99 times 9 = 99 times (10 - 1) = 99 times 10 - 99 times 1 ] [ = 990 - 99 = 891 ] 2. Next, we compare 891 with each of the given options to determine which is closest: - (A) 10000 is much larger than 891. - (B) 100 is much smaller than 891. - (C) 100000 is much larger than 891. - (D) 1000 is relatively close to 891. - (E) 10 is much smaller than 891.3. To confirm, the difference between 891 and 1000 is: [ 1000 - 891 = 109 ] Thus, the difference between 891 and 1000 is 109.4. Evaluating the difference with other options: - Difference between 891 and 10000: [ 10000 - 891 = 9109 ] - Difference between 891 and 100: [ 891 - 100 = 791 ] - Difference between 891 and 100000: [ 100000 - 891 approx 99109 ] - Difference between 891 and 10: [ 891 - 10 = 881 ]5. Comparing these differences, 109 (the difference to 1000) is the smallest. Thus, 1000 is the closest integer to (99 times 9). Conclusion:[boxed{text{D}}]

🧠:Okay, so I need to prove that there exist positive irrational numbers a and b such that a^b is a natural number. Hmm, interesting. Let me think about how to approach this. First, I know that rational numbers are fractions of integers, and irrational numbers can't be expressed that way. So, I need to find two numbers that are both irrational, but when one is raised to the power of the other, the result is an integer. Maybe I can start with a known example or think of some famous irrational numbers. Like sqrt(2) is irrational, right? And e is irrational, and pi is irrational. Maybe I can use combinations of these? Wait, what if I take a number like sqrt(2) and raise it to another irrational power? For example, sqrt(2)^sqrt(2) is irrational? I think that's actually unknown or maybe it's transcendental? Wait, no, actually, the Gelfond-Schneider theorem tells us that if a is algebraic (but not 0 or 1) and b is irrational algebraic, then a^b is transcendental. So sqrt(2) is algebraic, and sqrt(2) is irrational algebraic. So sqrt(2)^sqrt(2) is transcendental, which is irrational. But how does that help me here?Wait, maybe instead of taking both a and b as sqrt(2), I need to find a different pair. Let me think. Suppose I want a^b to be an integer, say 2. So, can I find irrational a and b such that a^b = 2?Let's see. Let's take a = 2 and b = 1. But both are rational. Not helpful. If I take a = 2^(1/b), then maybe. If I can choose b such that 1/b is irrational. Wait, but then a would be 2^(1/b). If b is irrational, then 1/b is irrational too. But then is 2^(1/b) irrational?Suppose b is irrational. Let's take b = sqrt(2). Then a = 2^(1/sqrt(2)). Is that irrational? Probably, but how do I prove it? Hmm. Alternatively, maybe use logarithms.Let me think of another approach. Suppose we let a = sqrt(2)^sqrt(2), which we know from Gelfond-Schneider is transcendental, hence irrational. Then let b = sqrt(2). Then a^b = (sqrt(2)^sqrt(2))^sqrt(2) = sqrt(2)^(sqrt(2)*sqrt(2)) = sqrt(2)^2 = 2, which is natural. But wait, is a = sqrt(2)^sqrt(2) irrational? Yes, as per Gelfond-Schneider, since sqrt(2) is algebraic not 0 or 1, and sqrt(2) is irrational algebraic, so a is transcendental, hence irrational. And b = sqrt(2) is irrational. So that works. Therefore, a and b are both irrational, and a^b = 2, which is natural.But wait, is that correct? Let me check. So if a = sqrt(2)^sqrt(2), which is transcendental, so definitely irrational. Then b = sqrt(2), which is irrational. Then a^b = (sqrt(2)^sqrt(2))^sqrt(2) = sqrt(2)^(sqrt(2)*sqrt(2)) = sqrt(2)^2 = 2. Yep, that seems to work. But hold on, do we actually know that sqrt(2)^sqrt(2) is irrational? The Gelfond-Schneider theorem tells us it's transcendental, which implies it's irrational. So yes. So that pair works. But wait, the problem is asking to prove existence, so even if we can't compute the exact values, as long as such numbers exist, we can use them. Alternatively, another example: take a = sqrt(2) and b = 2*log_2(3). Wait, let's see. If a = sqrt(2), which is irrational, and b = 2*log_2(3). If log_2(3) is irrational, then b is irrational. Let me check. log_2(3) is irrational because 3 is not a power of 2. Yes, so log_2(3) is irrational, so 2*log_2(3) is also irrational. Then a^b = (sqrt(2))^{2*log_2(3)} = (2^{1/2})^{2*log_2(3)} = 2^{(1/2)*2*log_2(3)} = 2^{log_2(3)} = 3, which is natural. So here, a = sqrt(2) is irrational, b = 2*log_2(3) is irrational, and a^b = 3. So that's another example. But wait, is log_2(3) irrational? Yes, because if log_2(3) = p/q for integers p and q, then 2^{p/q} = 3, which implies 2^p = 3^q. But 2 and 3 are primes, so this equation can't hold unless p = q = 0, which isn't possible. Hence, log_2(3) is irrational. So this example works too. So here, a is sqrt(2) (irrational), b is 2 log_2(3) (irrational), and a^b = 3. So that's another pair. So either of these examples would work. But in the first example, using a = sqrt(2)^sqrt(2) and b = sqrt(2), we can also get a^b = 2. So perhaps either of these examples can be used. But the problem is just to prove existence, so even one example suffices. Wait, but the first example uses a = sqrt(2)^sqrt(2), which is known to be transcendental, but do we need to prove that it's irrational? Since transcendental numbers are irrational, that's sufficient. Alternatively, the second example is more straightforward because we can directly verify the irrationality of a and b. Let me check the second example again. a = sqrt(2), which is irrational. b = 2 log_2(3). Since log_2(3) is irrational, multiplying by 2 (a rational number, non-zero) preserves irrationality. Because if 2 log_2(3) were rational, then log_2(3) would be rational, which it's not. Hence, b is irrational. Then a^b = sqrt(2)^{2 log_2(3)} = (2^{1/2})^{2 log_2(3)} = 2^{(1/2 * 2) log_2(3)} = 2^{log_2(3)} = 3. So that's a valid example. So both a and b are irrational, and their exponent is 3. Therefore, such numbers exist. Alternatively, another way to think about it is using Euler's formula with e and ln, but maybe that's more complicated. For example, take a = e and b = ln(n), where n is a natural number. But ln(n) is irrational for integers n > 1 except for powers of e, which don't exist since e is transcendental. Wait, ln(n) is irrational for integers n that are not powers of e. But e itself is transcendental, so there are no integer powers of e except e^1, e^2, etc., which are not integers. Hence, ln(n) is irrational for any integer n > 0 except n = 1, but ln(1) = 0, which is rational. So if we take a = e, which is irrational, and b = ln(2), which is irrational, then a^b = e^{ln(2)} = 2. But wait, in this case, a = e is irrational, b = ln(2) is irrational, and a^b = 2. So that's another example. So here's another pair: a = e, b = ln(2). Both are irrational, and e^{ln(2)} = 2. But wait, ln(2) is irrational, right? Yes, because if ln(2) = p/q, then e^{p/q} = 2, so e^p = 2^q. But e is transcendental, so it can't be equal to 2^{q/p}, which is algebraic. Hence, no such integers p and q exist, so ln(2) is irrational. Therefore, this gives another valid example. So there are multiple examples. Therefore, the existence is proven by example. But the problem says "positive irrational numbers", so all these examples are positive. Therefore, the answer is yes, such numbers exist, and examples include a = sqrt(2), b = 2 log_2(3); or a = e, b = ln(2); or a = sqrt(2)^sqrt(2), b = sqrt(2). So since we just need to prove existence, providing one explicit example is sufficient. Alternatively, if we don't want to rely on transcendental numbers, we can use the first example with a = sqrt(2) and b = 2 log_2(3). Because sqrt(2) is algebraic, and log_2(3) is irrational. But in any case, the key idea is to take a number, raise it to an irrational power to get an integer. Since logarithms can relate exponents to results. Another way: suppose I want a^b = n, where n is natural. Then, take a = n^{1/b}. If we can choose b such that 1/b is irrational, then a would be n^{1/b}. If b is irrational, then 1/b is irrational, so n^{1/b} might be irrational. But how to ensure that? But perhaps if we take b = log_n(k), where k is some number. Wait, this might not be straightforward. Alternatively, using the Gelfond-Schneider theorem, which says that if a and b are algebraic, a not 0 or 1, and b irrational, then a^b is transcendental. So, if we pick a = sqrt(2) (algebraic), b = sqrt(2) (algebraic irrational), then a^b is transcendental, which is irrational. Then, raising that to another sqrt(2) gives us 2. But maybe this is the same as the first example. Alternatively, perhaps even simpler: take a = 2^{sqrt(2)} and b = 1/sqrt(2). Then, a^b = (2^{sqrt(2)})^{1/sqrt(2)} = 2^{sqrt(2)/sqrt(2)} = 2^1 = 2. Now, is a = 2^{sqrt(2)} irrational? Yes, because if it were rational, then 2^{sqrt(2)} = p/q for integers p and q. But sqrt(2) is irrational, so 2^{sqrt(2)} is transcendental by Gelfond-Schneider, hence irrational. And b = 1/sqrt(2) is irrational. Therefore, both a and b are irrational, and a^b = 2. So this is another valid example. So this seems to be a more straightforward example. Let me check again:a = 2^{sqrt(2)}, which is irrational (even transcendental), and b = 1/sqrt(2), which is irrational. Then, a^b = 2^{sqrt(2) * (1/sqrt(2))} = 2^1 = 2. Yes, that works. So this might be a simpler example. So in this case, both a and b are explicitly constructed irrational numbers, and their exponentiation yields 2. Therefore, such numbers do exist. So, summarizing, there are multiple ways to construct such pairs. The key idea is to take a known irrational number, manipulate its exponent in such a way that the result is an integer. Using properties of exponents and logarithms, and relying on the fact that certain exponentials yield transcendental (hence irrational) numbers. Another approach is to consider numbers like sqrt(n) and logarithms. For example, if we take a = n^{1/b} where b is irrational, then if 1/b is irrational, a might be irrational. Then a^b = n. But this requires that a = n^{1/b} is irrational. Alternatively, using Euler's identity, but that's more complex. In any case, since the problem only asks for existence, providing a single explicit example suffices. Therefore, the answer is yes, and examples include a = sqrt(2), b = 2 log_2(3), or a = 2^{sqrt(2)}, b = 1/sqrt(2), etc.

✅:To prove that there exist positive irrational numbers (a) and (b) such that (a^{b}) is a natural number, we can construct explicit examples using properties of exponents and logarithms.Example 1: Let (a = sqrt{2}) and (b = 2 log_2 3). - Irrationality of (a): (sqrt{2}) is irrational. - Irrationality of (b): (log_2 3) is irrational (since (3) is not a power of (2)), and multiplying by (2) preserves irrationality. - Result: [ a^{b} = (sqrt{2})^{2 log_2 3} = 2^{frac{1}{2} cdot 2 log_2 3} = 2^{log_2 3} = 3. ] Here, (a^b = 3 in mathbb{N}).Example 2: Let (a = 2^{sqrt{2}}) and (b = frac{1}{sqrt{2}}). - Irrationality of (a): By the Gelfond-Schneider theorem, (2^{sqrt{2}}) is transcendental (hence irrational). - Irrationality of (b): (frac{1}{sqrt{2}}) is irrational. - Result: [ a^{b} = left(2^{sqrt{2}}right)^{frac{1}{sqrt{2}}} = 2^{sqrt{2} cdot frac{1}{sqrt{2}}} = 2^{1} = 2. ] Here, (a^b = 2 in mathbb{N}).Conclusion: These examples demonstrate that such pairs ((a, b)) exist. Therefore, there are positive irrational numbers (a) and (b) for which (a^{b}) is a natural number.(boxed{text{Such numbers exist, e.g., } a = sqrt{2}, b = 2 log_2 3 text{ or } a = 2^{sqrt{2}}, b = frac{1}{sqrt{2}}.})

🔑:1. Initial Setup: We need to prove that there exist positive irrational numbers (a) and (b) such that (a^b) is a natural number.2. Choosing Numbers: We choose the following numbers: [ a = sqrt{2} ] and [ b = log_{sqrt{2}}(3) ]3. Evaluate (a^b): We want to check if (a^b) is a natural number: [ a^b = left(sqrt{2}right)^{log_{sqrt{2}}(3)} ]4. Simplify the Expression: Notice that (sqrt{2} = 2^{1/2}). Therefore, [ a^b = left(2^{1/2}right)^{log_{sqrt{2}}(3)} ]5. Change of Base Formula: Using the logarithm change of base formula, we know: [ log_{sqrt{2}}(3) = frac{log_{2}(3)}{log_{2}(sqrt{2})} ] Since (log_{2}(sqrt{2}) = 1/2): [ log_{sqrt{2}}(3) = frac{log_{2}(3)}{1/2} = 2 log_{2}(3) ]6. Substitute Back: So: [ a^b = left(2^{1/2}right)^{2 log_{2}(3)} ] [ a^b = 2^{left(frac{1}{2} cdot 2 log_{2}(3)right)} = 2^{log_{2}(3)} ]7. Exponent Properties: Using the property of exponents (a^{log_{a}(x)} = x), we get: [ 2^{log_{2}(3)} = 3 ] Thus, we have shown that: [ a^b = 3 ]8. Verify Irrationality: Next, we verify that both (a) and (b) are irrational numbers. begin{itemize} item (a = sqrt{2}) is known to be irrational. item We need to show that (b = log_{sqrt{2}}(3)) is also irrational. Assume, for contradiction, (b) is rational: [ b = log_{sqrt{2}}(3) = frac{p}{q} text{ where } p, q in mathbb{N} ] This implies: [ sqrt{2}^{p/q} = 3 ] Raising both sides to the power (q): [ left(2^{1/2}right)^p = 3^q implies 2^{p/2} = 3^q ] But (2^{p/2}) is a power of 2, and (3^q) is a power of 3. Since 2 and 3 are distinct primes, (2^{p/2} = 3^q) is impossible for any positive integers (p) and (q). Thus, (b) cannot be rational. end{itemize}Therefore, both (a = sqrt{2}) and (b = log_{sqrt{2}}(3)) are positive irrational numbers such that (a^b = 3), which is a natural number.[boxed{a = sqrt{2}, b = log_{sqrt{2}}(3)}]

🧠:Okay, let's see. I need to find the number of points P inside a unit circle such that the sum of the squares of the distances from P to the endpoints of a diameter is 3. The options are 0, 1, 2, 4, or infinite. Hmm, let's break this down step by step.First, let's visualize the problem. There's a circle with radius 1. A diameter is a straight line passing through the center, so the endpoints of the diameter are two points on the circle that are directly opposite each other. Let me denote these endpoints as A and B. The points P are somewhere inside the circle, and for each P, we calculate the square of the distance from P to A and the square of the distance from P to B, then add them together. The sum should equal 3. We need to find how many such points P exist.Let me start by setting up a coordinate system to make this easier. Let's place the circle with its center at the origin (0,0) on a coordinate plane. Without loss of generality, we can take the diameter along the x-axis. So, point A is (1, 0) and point B is (-1, 0). Then, any point P inside the circle can be represented as (x, y), where x² + y² ≤ 1.The distance from P to A is the distance between (x, y) and (1, 0). Using the distance formula, that's sqrt[(x - 1)² + (y - 0)²]. Similarly, the distance from P to B is sqrt[(x + 1)² + y²]. The problem states that the sum of the squares of these distances is 3. Since squaring the distances will eliminate the square roots, let's compute that.So, the square of the distance from P to A is (x - 1)² + y², and the square of the distance from P to B is (x + 1)² + y². Adding these together:Sum = (x - 1)² + y² + (x + 1)² + y²Let me expand these terms:First, expand (x - 1)²: x² - 2x + 1Then, expand (x + 1)²: x² + 2x + 1So substituting back into the sum:Sum = (x² - 2x + 1 + y²) + (x² + 2x + 1 + y²)Combine like terms:x² + x² = 2x²-2x + 2x = 0 (they cancel out)1 + 1 = 2y² + y² = 2y²So the sum simplifies to:2x² + 2y² + 2We are told this sum equals 3:2x² + 2y² + 2 = 3Subtract 2 from both sides:2x² + 2y² = 1Divide both sides by 2:x² + y² = 0.5Hmm, interesting. So the set of points P that satisfy the given condition lie on a circle centered at the origin with radius sqrt(0.5) ≈ 0.707. But wait, the original condition says that P must be inside the unit circle. The circle with radius sqrt(0.5) is entirely within the unit circle because sqrt(0.5) is less than 1. So, the points P that satisfy the condition lie on this smaller circle. Therefore, all the points on this smaller circle would satisfy the equation x² + y² = 0.5. Since a circle consists of infinitely many points, one might think the answer is infinite. But let me double-check because the options include (E) Infinite, but maybe there's a restriction I'm missing.Wait, the problem states that the points P are within the region enclosed by the circle of radius 1. Wait, does "within" include the boundary or not? Hmm, in some contexts, "within" can mean strictly inside, but in others, it can include the boundary. The original circle here has radius 1, so if the problem says points within the region, maybe including the boundary. However, the points P that satisfy x² + y² = 0.5 are strictly inside the unit circle since 0.5 < 1. Therefore, even if "within" excludes the boundary, these points are still inside. So, if the equation x² + y² = 0.5 is a full circle inside the unit circle, then there are infinitely many such points. But the answer options are (A) 0, (B)1, (C)2, (D)4, (E)Infinite. So, (E) would be the answer. But wait, the answer is (E) then? Wait, but let me check again because maybe there's a mistake in my calculation.Wait, let's go back step by step. The problem says the sum of the squares of the distances to the endpoints of a diameter is 3. We set up coordinates with A(1,0) and B(-1,0). Then the sum of squares is:[(x - 1)^2 + y^2] + [(x + 1)^2 + y^2] = 3Expanding that, as I did before, gives 2x² + 2y² + 2 = 3, leading to x² + y² = 0.5. So all points on the circle with radius sqrt(0.5) centered at the origin satisfy this. Since this circle is entirely within the unit circle, then all these points are indeed within the region. Therefore, there are infinitely many such points. So the answer should be (E) Infinite.But wait, let me check the problem statement again. It says "points P are within a region enclosed by a circle of radius 1". If "within" is strictly inside, not including the boundary, but the circle x² + y² = 0.5 is entirely inside, so even if the original circle's boundary is excluded, the points on x² + y² = 0.5 are still within. Therefore, the number of points is infinite. So, is the answer (E)?But wait, the answer options given here are from the 20th Annual U.S. High School Mathematics Examination (1969), which is an older version of the AMC 12. Let me see if I recall correctly that sometimes in such problems, even if it seems like a circle, there might be constraints. Wait, but if you have a circle of radius sqrt(0.5) inside the unit circle, that's a valid locus of points. So, the intersection is the entire circle, which has infinitely many points. So the answer should be (E).But hold on, the answer options in the original problem are (A) 0, (B)1, (C)2, (D)4, (E)Infinite. So the answer is (E). However, maybe there's a mistake in my reasoning. Let me check once again.Alternative approach: Let me consider coordinates again. Let me parametrize point P as (x, y). Then, sum of squares of distances to A(1,0) and B(-1,0):[(x - 1)^2 + y^2] + [(x + 1)^2 + y^2] = 3Expanding both:(x² - 2x +1 + y²) + (x² + 2x +1 + y²) = 3Combine like terms: 2x² + 2y² + 2 = 3Subtract 2: 2x² + 2y² = 1Divide by 2: x² + y² = 1/2Yes, exactly. So this is a circle with radius sqrt(1/2) ≈ 0.707, centered at the origin. Since the original circle has radius 1, this circle is entirely contained within it. Therefore, all points on this circle are within the unit circle. Therefore, there are infinitely many such points, so answer is (E). But let me check if maybe the problem is in 3D? No, the problem mentions a region enclosed by a circle, so it's in 2D. So, that should be okay.Wait, but the answer given in some sources might be different. Wait, perhaps I made a mistake in the algebra. Let me check the expansion again.First distance squared: (x - 1)^2 + y^2 = x² - 2x + 1 + y²Second distance squared: (x + 1)^2 + y^2 = x² + 2x + 1 + y²Adding them: (x² - 2x + 1 + y²) + (x² + 2x + 1 + y²) = 2x² + 2y² + 2Yes, that's correct. Then 2x² + 2y² + 2 = 3 → 2x² + 2y² = 1 → x² + y² = 1/2. Correct.Therefore, the set of points P lies on a circle of radius 1/sqrt(2). Since this circle is entirely within the unit circle, all its points are inside. Therefore, there are infinitely many points P satisfying the condition. Therefore, the answer is (E).But wait, the answer options here are from 1969. Maybe there is a trick here. Wait, maybe the problem says "points within a region enclosed by a circle of radius 1", but the locus x² + y² = 1/2 is a circle, which is a 1-dimensional figure, so technically, it has infinitely many points. Therefore, unless there's a misinterpretation, the answer should be (E). But the options given include (E)Infinite, which would be correct.Alternatively, maybe the problem is from an old exam where the answer is different. Wait, let me check if maybe the original problem had a different diameter. Wait, in the problem statement, it's a diameter, so the endpoints are definitely two points opposite each other. The calculation seems straightforward. Let me also think of specific points. For example, the center of the circle is (0,0). Let's check the sum of squares of distances from the center to A and B.Distance from center to A is 1, so square is 1. Similarly, distance from center to B is also 1, so sum is 1 + 1 = 2. But the problem states the sum should be 3, so the center does not satisfy the condition. What about a point on the circle x² + y² = 1/2, say (1/√2, 0). Then distance to A is |1/√2 - 1|, squared is (1 - 1/√2)^2. Distance to B is |1/√2 + 1|, squared is (1 + 1/√2)^2. Adding them:(1 - 1/√2)^2 + (1 + 1/√2)^2 = [1 - 2/√2 + 1/2] + [1 + 2/√2 + 1/2] = [1.5 - 2/√2] + [1.5 + 2/√2] = 3. So yes, that works. Therefore, (1/√2, 0) is one such point, and similarly, all points on that circle work. Therefore, infinitely many. So the answer is (E).But let me check again the problem statement to see if there's a misinterpretation. It says "points P are within a region enclosed by a circle of radius 1". So, within, not on. But the circle x² + y² = 1/2 is entirely within the unit circle. However, in topology, "within" can sometimes mean strictly inside, but in geometry problems, unless specified otherwise, I think points on the boundary (of the region) are considered. However, in this case, the points in question are on the circle x² + y² = 1/2, which is strictly inside the unit circle. Therefore, even if "within" excludes the boundary of the unit circle, these points are still within. Therefore, the set of solutions is the circle x² + y² = 1/2, which has infinitely many points. Therefore, the answer is (E)Infinite.But wait, the answer given in the options is (E). However, maybe the original problem had a different configuration. Let me check once more.Alternatively, maybe there's a mistake in my calculation. Let me recast the problem using vectors. Let me consider points A and B as vectors (1,0) and (-1,0). Then for any point P with coordinates (x,y), the square of the distance PA is |P - A|² = (x - 1)^2 + y^2, similarly for PB. Then PA² + PB² = (x - 1)^2 + y^2 + (x + 1)^2 + y^2 = 2x² + 2y² + 2, which equals 3. Therefore, x² + y² = 0.5. That seems correct.Alternatively, maybe there's a property from geometry that can be used here. The sum of the squares of the distances from any point to two fixed points. I recall that in plane geometry, the sum of the squares of the distances from a point to two fixed points can be expressed in terms of the coordinates. But in this case, we have a specific sum.Alternatively, using coordinate geometry, as above, which leads to the same conclusion. So, given that the equation reduces to a circle, which lies entirely within the unit circle, then the number of points is infinite. Therefore, the answer should be (E). However, the options given in the problem are (A)0, (B)1, (C)2, (D)4, (E)Infinite. So, unless there is a mistake in the problem statement, or in my reasoning, the correct answer is (E).But wait, perhaps I made a mistake in the problem statement. Let me check again.The problem says: "the sum of the squares of the distances from these P points to the endpoints of a diameter is 3". So each such P must satisfy PA² + PB² = 3, where A and B are endpoints of a diameter of the unit circle. Then, through calculation, we derived that the locus is the circle x² + y² = 0.5. Thus, the points P lie on this circle. Since this circle is inside the unit circle, all points on it are within the region. Therefore, there are infinitely many such points.Therefore, the answer is (E) Infinite.But the original answer choices are from 1969. Maybe at that time, the answer was different. Wait, maybe the problem is in 3D? No, it's enclosed by a circle, which is 2D. Hmm. Alternatively, maybe the diameter is not along the x-axis? Wait, but in the problem, it's just a diameter; the choice of coordinate system is arbitrary. So regardless of the diameter's orientation, the result should be the same. Because the circle is symmetric, so the equation would be similar for any diameter. Therefore, the answer remains the same.Alternatively, maybe I'm misapplying the term "within". If "within" is meant to be strictly inside, not including the boundary, but the circle x² + y² = 0.5 is strictly inside the unit circle. Therefore, points on x² + y² = 0.5 are within the unit circle. So even if the original circle's boundary is excluded, these points are still included. So infinite.Alternatively, maybe the problem is a trick question. If you think about the sum of squares of distances from P to A and B. For any point P, PA² + PB². Let me recall a theorem: For any point P and a segment AB with midpoint M, PA² + PB² = 2PM² + (AB²)/2. Wait, yes, that's the formula from coordinate geometry. So, in this case, AB is the diameter of the unit circle, so length AB is 2. The midpoint M is the center of the circle, which is the origin. Therefore, PA² + PB² = 2PM² + (AB²)/2. AB is 2, so AB² is 4, so (AB²)/2 is 2. Therefore, PA² + PB² = 2PM² + 2. According to the problem, this sum is 3. So:2PM² + 2 = 3 → 2PM² = 1 → PM² = 0.5 → PM = sqrt(0.5). Therefore, the set of points P is the set of points at distance sqrt(0.5) from M, the center. So, a circle with radius sqrt(0.5) around the center. Which is exactly what we derived earlier. Therefore, confirming that it's a circle, hence infinitely many points.Therefore, the answer should be (E)Infinite.But perhaps I need to check if sqrt(0.5) is indeed less than 1. Since sqrt(0.5) ≈ 0.707, which is less than 1, so yes, the circle is entirely inside the unit circle. Therefore, all points on this circle are within the unit circle, so n is infinite. Hence, option (E).I think that's thorough. Unless there is a trick or a misinterpretation, but I can't see any. Therefore, the answer is (E).Final Answerboxed{E}

✅:To determine the number of points ( P ) within a unit circle such that the sum of the squares of the distances from ( P ) to the endpoints of a diameter is 3, we start by setting up a coordinate system. Let the endpoints of the diameter be ( A(1, 0) ) and ( B(-1, 0) ). For any point ( P(x, y) ) inside the unit circle, the distances to ( A ) and ( B ) are given by:[PA = sqrt{(x - 1)^2 + y^2} quad text{and} quad PB = sqrt{(x + 1)^2 + y^2}]The sum of the squares of these distances is:[PA^2 + PB^2 = (x - 1)^2 + y^2 + (x + 1)^2 + y^2]Expanding and simplifying the expression:[(x - 1)^2 + y^2 + (x + 1)^2 + y^2 = (x^2 - 2x + 1 + y^2) + (x^2 + 2x + 1 + y^2)][= 2x^2 + 2y^2 + 2]Given that this sum equals 3:[2x^2 + 2y^2 + 2 = 3][2x^2 + 2y^2 = 1][x^2 + y^2 = frac{1}{2}]This equation represents a circle centered at the origin with radius (sqrt{frac{1}{2}} = frac{sqrt{2}}{2}). Since this radius is less than 1, the circle is entirely contained within the unit circle. Therefore, all points on this circle are within the unit circle, leading to infinitely many such points ( P ).Thus, the number of such points ( P ) is (boxed{E}).

🔑:1. We start with a circle of radius 1, centered at the origin in the Cartesian coordinate system. The equation of this circle is [ x^2 + y^2 = 1. ]2. Given that A and B are the endpoints of a diameter, their coordinates will be (-1,0) and (1,0) respectively.3. Let P = (x, y) be any point within or on the circle. We need to consider the sum of the squares of the distances from P to A and P to B. This can be written as: [ left[(x+1)^2 + y^2right] + left[(x-1)^2 + y^2right] = 3. ]4. We can expand and simplify the above expression: [ left[(x+1)^2 + y^2right] + left[(x-1)^2 + y^2right] = (x^2 + 2x + 1 + y^2) + (x^2 - 2x + 1 + y^2). ] Combine like terms: [ = x^2 + 2x + 1 + y^2 + x^2 - 2x + 1 + y^2 = 2x^2 + 2y^2 + 2. ]5. Given the problem statement, we set the simplified expression equal to 3: [ 2x^2 + 2y^2 + 2 = 3. ] Subtract 2 from both sides: [ 2x^2 + 2y^2 = 1. ] Divide everything by 2: [ x^2 + y^2 = frac{1}{2}. ]6. The equation x^2 + y^2 = frac{1}{2} represents a circle centered at the origin with radius (sqrt{frac{1}{2}} = frac{sqrt{2}}{2}), which is smaller than the original circle of radius 1.7. As any point P that satisfies x^2 + y^2 = frac{1}{2} also lies within the larger circle of radius 1, we can conclude that this secondary, smaller circle lies entirely within the original circle.8. Hence, there are infinitely many such points P that meet the required condition.Conclusion:[boxed{text{E}}]

🧠:Okay, so I need to find the minimum possible denominator of the sum of two irreducible fractions where each fraction has a denominator of 600 and 700 respectively. Hmm, let me start by recalling some basics about fractions and denominators.First, when you add two fractions, the denominator of the sum is the least common multiple (LCM) of their denominators, provided they are in their simplest forms. But wait, the problem says the fractions themselves are irreducible. So, each fraction a/600 and b/700 must already be in their simplest form, meaning that the numerators a and b don't share any common factors with their denominators 600 and 700, respectively.Right, so since 600 and 700 are the denominators, and the fractions are irreducible, the numerators must be coprime with 600 and 700. Now, when adding a/600 + b/700, the common denominator would be LCM(600, 700). Let me compute that.First, let's factorize 600 and 700 into primes.600 = 2^3 * 3 * 5^2700 = 2^2 * 5^2 * 7So the LCM is the product of the highest powers of all primes present:LCM(600, 700) = 2^3 * 3 * 5^2 * 7 = 8 * 3 * 25 * 7Calculating that: 8*3=24, 24*25=600, 600*7=4200. So LCM is 4200. Therefore, the denominator of the sum would be 4200 if we just add them directly. But the question is asking for the minimum possible denominator after simplifying the sum. So maybe 4200 isn't the minimum? Because if the numerator and denominator of the sum have a common factor, we can reduce the fraction further.Therefore, the key is to find two fractions a/600 and b/700 (both irreducible) such that when added together, the resulting fraction (a/600 + b/700) can be simplified as much as possible, leading to the smallest possible denominator.So, let me formalize this. Let's express the sum:a/600 + b/700 = (7a + 6b)/4200Because 4200 is the LCM of 600 and 700. So the numerator is 7a + 6b, and the denominator is 4200. Then, the fraction (7a + 6b)/4200 can be simplified by dividing both numerator and denominator by their greatest common divisor (GCD). So, the denominator after simplification would be 4200 divided by GCD(7a + 6b, 4200). Therefore, to minimize the denominator, we need to maximize the GCD of the numerator (7a + 6b) and 4200.Therefore, the problem reduces to finding the maximum possible value of d = GCD(7a + 6b, 4200), where a is coprime with 600, and b is coprime with 700. Then, the minimal denominator would be 4200 / d.So, my goal is to maximize d, such that d divides 4200, and there exist integers a and b with a coprime to 600, b coprime to 700, and d divides 7a + 6b.Hmm, okay. Let's note that 4200 factors into primes as 2^3 * 3 * 5^2 * 7, same as the LCM.So, to find the maximum possible d, we need to find the largest divisor of 4200 such that 7a + 6b ≡ 0 mod d, given the coprimality conditions on a and b.Alternatively, perhaps we can think of this as solving the congruence 7a + 6b ≡ 0 mod d, and we need to find the largest d dividing 4200 for which such a and b exist with a coprime to 600 and b coprime to 700.But how do we approach this? Maybe we can analyze the possible common divisors d of 7a + 6b and 4200.Let me consider the prime factors of 4200: 2, 3, 5, 7.To maximize d, we need to include the highest possible powers of these primes in d, provided that for each prime power in d, we can have 7a + 6b ≡ 0 mod that prime power, given the constraints on a and b.So, let's check each prime factor:1. Prime 2: 4200 has 2^3. Let's see if d can have 2^3.2. Prime 3: 4200 has 3^1. Can d include 3?3. Prime 5: 4200 has 5^2. Can d include 5^2?4. Prime 7: 4200 has 7^1. Can d include 7?So, we need to check for each prime power whether it can divide 7a + 6b given the coprimality conditions on a and b. Let's handle each prime separately.Starting with prime 2:We need to see what possible exponents of 2 can divide 7a + 6b. Let's consider modulus 2, 4, 8.First, modulo 2:7a + 6b ≡ 1*a + 0*b ≡ a mod 2.But since a is coprime to 600, which is divisible by 2, a must be odd. Therefore, a ≡ 1 mod 2. Thus, 7a + 6b ≡ 1 mod 2. So, the sum is 1 mod 2, which means 2 does not divide the numerator. Therefore, d cannot have any factors of 2. Wait, this is a key point. Because a must be coprime to 600, and since 600 is divisible by 2, 3, and 5, a cannot be even, cannot be a multiple of 3 or 5. So a is odd, not divisible by 3 or 5. Similarly, b must be coprime to 700, which is divisible by 2, 5, 7. So b must be not divisible by 2, 5, or 7.Therefore, a is odd, not divisible by 3 or 5. b is odd (since 700 is divisible by 2, so b must be coprime to 2, hence odd), not divisible by 5 or 7.So, 7a is 7 times an odd number, which is odd (since 7 is odd and a is odd), and 6b is 6 times an odd number, which is even. Therefore, 7a is odd, 6b is even, so their sum is odd + even = odd. Therefore, 7a + 6b is odd, so it cannot be divisible by 2. Therefore, the numerator is odd, so GCD(7a + 6b, 4200) must be odd. Therefore, the factor 2^3 in 4200 is irrelevant for the GCD. So, d cannot have any factors of 2. Therefore, the maximum possible d is a divisor of 4200 / 8 = 525, which is 3 * 5^2 * 7. So, d must divide 525.Now, moving to prime 3:We need to check if 3 can divide 7a + 6b. Let's compute 7a + 6b mod 3.7 ≡ 1 mod 3, 6 ≡ 0 mod 3. Therefore, 7a + 6b ≡ a mod 3. So, 7a + 6b ≡ a mod 3. Since a is coprime to 600, which is divisible by 3, a cannot be divisible by 3, so a ≡ 1 or 2 mod 3. Therefore, 7a + 6b ≡ 1 or 2 mod 3. Therefore, the sum cannot be divisible by 3. Therefore, 3 does not divide the numerator. So, d cannot have 3 as a factor. Therefore, d must divide 525 / 3 = 175 = 5^2 * 7.Moving on to prime 5:We need to check if 5 divides 7a + 6b. Let's compute 7a + 6b mod 5.7 ≡ 2 mod 5, 6 ≡ 1 mod 5. Therefore, 7a + 6b ≡ 2a + b mod 5.We need to check if 2a + b ≡ 0 mod 5. But a is coprime to 600, so a is not divisible by 5. Similarly, b is coprime to 700, so b is not divisible by 5. Therefore, a can be 1, 2, 3, or 4 mod 5, and similarly for b.So, can we choose a and b such that 2a + b ≡ 0 mod 5?Yes. For example, let's set a ≡ 1 mod 5. Then, 2*1 + b ≡ 0 mod 5 ⇒ b ≡ -2 mod 5 ⇒ b ≡ 3 mod 5. Since b must not be divisible by 5, but can be 3 mod 5. Similarly, other choices: a ≡ 2 mod 5 ⇒ 2*2 = 4 ⇒ b ≡ -4 ≡ 1 mod 5. So, as long as we can choose a and b such that their congruence classes mod 5 satisfy 2a + b ≡ 0 mod 5, then 5 divides the sum. But is that possible?Yes. Since a and b are not divisible by 5, but their residues mod 5 can be adjusted. Since a can be 1, 2, 3, 4 mod 5, and b can be 1, 2, 3, 4 mod 5, we can always solve 2a + b ≡ 0 mod 5. For each a, there is a b that works, and for each b, there is an a. For example:If a ≡ 1 mod 5, then b ≡ -2*1 ≡ 3 mod 5.If a ≡ 2 mod 5, then b ≡ -2*2 ≡ -4 ≡ 1 mod 5.If a ≡ 3 mod 5, then b ≡ -2*3 ≡ -6 ≡ -1 ≡ 4 mod 5.If a ≡ 4 mod 5, then b ≡ -2*4 ≡ -8 ≡ -3 ≡ 2 mod 5.Therefore, for each a not divisible by 5, there exists a b not divisible by 5 such that 2a + b ≡ 0 mod 5. Therefore, 5 divides 7a + 6b. Therefore, 5 is a factor of the GCD.But can we get higher powers of 5? Let's check mod 25.Compute 7a + 6b mod 25. Let's see:7a + 6b ≡ 7a + 6b mod 25.We need to see if 7a + 6b ≡ 0 mod 25, given that a ≡ not 0 mod 5 and b ≡ not 0 mod 5.But is it possible for 7a + 6b ≡ 0 mod 25?This is more complex. Let's see. Let's assume that a and b are chosen such that 2a + b ≡ 0 mod 5 (as before), so 2a + b = 5k for some integer k. Then, we need to see if 7a + 6b ≡ 0 mod 25.Express b in terms of a: from 2a + b ≡ 0 mod 5, we have b ≡ -2a mod 5. Let's write b = -2a + 5m for some integer m.Substitute into 7a + 6b:7a + 6*(-2a + 5m) = 7a -12a + 30m = -5a + 30m.So, 7a + 6b = -5a + 30m = 5*(-a + 6m). So, it's divisible by 5, but we need it to be divisible by 25. Therefore, -a + 6m ≡ 0 mod 5 ⇒ -a + 6m ≡ 0 mod 5 ⇒ a ≡ 6m mod 5 ⇒ a ≡ m mod 5 (since 6 ≡ 1 mod 5). Therefore, if we set m ≡ a mod 5, then this condition is satisfied. Let's let m = a + 5n for some integer n. Then, substituting back:7a + 6b = 5*(-a + 6m) = 5*(-a + 6(a + 5n)) = 5*(-a + 6a + 30n) = 5*(5a + 30n) = 25(a + 6n).Therefore, 7a + 6b is divisible by 25. Therefore, if we can choose a and b such that m is an integer (i.e., b = -2a + 5m = -2a + 5(a + 5n) = -2a +5a +25n = 3a +25n), then 7a + 6b =25(a +6n). Therefore, if we can choose a and b in such a way, then 25 divides the numerator.But we need to ensure that a is coprime to 600 and b is coprime to 700. Let's see. Let's pick a specific example.Let me choose a =1. Since a must be coprime to 600, which is true because 1 is coprime to everything. Then, b =3*1 +25n =3 +25n. Let's pick n=0: then b=3. Check if b=3 is coprime to 700: 700 factors into 2^2 *5^2 *7. 3 is coprime to 2,5,7, so yes, 3 is coprime to 700. Therefore, a=1, b=3 gives 7*1 +6*3=7 +18=25, which is divisible by 25. Therefore, in this case, GCD(25,4200)=25, so the denominator would be 4200/25=168. But wait, but maybe even higher?Wait, no. 25 is 5^2. The maximum power of 5 in 4200 is 5^2. So, we can get 5^2 as part of the GCD. Similarly, we need to check if 7 can be part of the GCD.Let's check prime 7:Compute 7a +6b mod 7.7a ≡0 mod7, 6b≡6b mod7. Therefore, 7a +6b ≡6b mod7. So, 6b ≡0 mod7 ⇒b≡0 mod7. But b is coprime to 700, which is divisible by 7, so b cannot be divisible by7. Therefore, 6b≡0 mod7 only if b≡0 mod7, which is impossible. Therefore, 7 cannot divide the numerator. Therefore, 7 is not a factor of GCD(7a +6b,4200). Therefore, d cannot include 7.Therefore, the prime factors of d can only be 5, with exponent 2. Wait, but we saw earlier that 7a +6b can be divisible by 25, but not by 7 or 2 or 3. So, the maximum possible d is 25. Therefore, the minimal denominator would be 4200 /25=168.But wait, let's confirm this. Let's take a=1, b=3. Then, the sum is (1/600 +3/700). Let's compute this:1/600 =7/4200, 3/700=18/4200. Adding them gives 25/4200=1/168. So, denominator is 168. Therefore, 168 is achievable.But is it possible to get a higher d? For example, 25* something? But we saw that 7 cannot divide d, 2 and 3 cannot divide d. So, the maximum d is 25, so minimal denominator is 168.But wait, let's check another example. Suppose a= 1, b=3: sum is 25/4200=1/168. Another example: a=2, then b=3*2 +25n=6 +25n. Let n=0, then b=6, but 6 is not coprime to 700 (since 6 is divisible by 2, but 700 is divisible by 2, so b must be coprime to 2. 6 is even, so invalid. So, need to pick n such that b is odd. Since 25n is even if n is even, odd if n is odd. So, 6 +25n: if n is odd, 25n is odd, so 6 + odd= odd. Therefore, if n=1, then b=6 +25=31. 31 is coprime to 700. Then, a=2 (coprime to 600? 2 is not coprime to 600, since 600 is divisible by 2. Wait, a must be coprime to 600. a=2 is not allowed. So, a must be coprime to 600, meaning a cannot be divisible by 2,3,5. Therefore, a must be 1,7,11,13, etc. So, my previous example with a=1 is valid. If a=7, then b=3*7 +25n=21 +25n. Let n=0, b=21. Check if 21 is coprime to 700: 21=3*7, but 700 is divisible by 7, so b=21 is divisible by 7, which is not allowed. So, invalid. So n=1: b=21+25=46, which is even, so not coprime to 2. n=2: b=21+50=71. 71 is coprime to 700. So, a=7, b=71. Then, 7a +6b=7*7 +6*71=49 +426=475. 475 divided by 25 is 19. So, 475/4200=19/168. Denominator is 168. So again, denominator 168.Wait, but in this case, the GCD is 25, same as before. So, the GCD is 25, hence denominator 168.Is there a way to get a higher GCD? For example, if we can have 7a +6b divisible by 25 and some other prime? But we saw that 2,3,7 cannot divide the numerator. So, the GCD can only be 5^2=25. Therefore, the minimal denominator is 4200/25=168.But let's confirm with another example. Suppose a=1, b=3: sum is 1/600 +3/700= (7 + 18)/4200=25/4200=1/168. Yes. Another example: a=11, which is coprime to 600. Then, b=3*11 +25n=33 +25n. Let's take n=0: b=33. Check if 33 is coprime to 700: 33=3*11. 700 is divisible by 2,5,7, so 33 is coprime. Then, 7*11 +6*33=77 +198=275. 275 divides by 25? 275/25=11. So, 275=25*11. Therefore, GCD(275,4200)=25, since 4200=25*168, and 168 and 11 are coprime. So denominator is 168. Same as before.Therefore, it seems that the maximum possible GCD is 25, leading to the minimal denominator of 168. Therefore, the answer is 168.But wait, let me check once more. Suppose we take a=1 and b=3, giving GCD 25. What if we can find a and b such that 7a +6b is divisible by a higher power? But 4200 only has 5^2. So even if 7a +6b were divisible by, say, 5^3=125, but since 4200 only has 5^2, the GCD cannot exceed 25. Therefore, the maximum possible GCD is 25.Therefore, the minimal denominator is 4200 /25=168.But just to be thorough, suppose someone thinks that maybe with different a and b, you can get a higher GCD. Let's test a=1, b=3: sum numerator=25. a=1, b=28: wait, but b=28 is even, which is not allowed. a=1, b=3+25=28: invalid. a=1, b=3+25*2=53: 53 is coprime to 700. Then, 7*1 +6*53=7 +318=325. 325=25*13. GCD(325,4200)=25, same as before. So still 25.Another case: a=1, b=3+25*4=103. 7*1 +6*103=7 +618=625. 625=5^4. But 4200 has only 5^2. So GCD(625,4200)=25. Therefore, even if numerator is divisible by higher power of 5, the GCD is limited by the denominator's 5^2. Therefore, 25 is the maximum.Therefore, the minimal denominator is 168.Hence, the final answer is boxed{168}.

✅:To find the minimum value for the denominator of the sum of two irreducible fractions with denominators 600 and 700, we start by noting that the sum of the fractions ( frac{a}{600} ) and ( frac{b}{700} ) (where ( a ) and ( b ) are coprime to 600 and 700 respectively) is given by:[frac{a}{600} + frac{b}{700} = frac{7a + 6b}{4200}]The denominator of this sum, after simplification, is ( frac{4200}{gcd(7a + 6b, 4200)} ). To minimize this denominator, we need to maximize ( gcd(7a + 6b, 4200) ).Prime Factorization and Constraints:- ( 4200 = 2^3 times 3 times 5^2 times 7 )- ( a ) must be coprime to 600 (i.e., not divisible by 2, 3, or 5)- ( b ) must be coprime to 700 (i.e., not divisible by 2, 5, or 7)Analysis of Prime Factors:1. Prime 2: ( 7a + 6b ) is odd (since ( a ) and ( b ) are odd), so 2 cannot be a factor.2. Prime 3: ( 7a + 6b equiv a mod 3 ). Since ( a ) is not divisible by 3, 3 cannot be a factor.3. Prime 5: ( 7a + 6b equiv 2a + b mod 5 ). We can choose ( a ) and ( b ) such that ( 2a + b equiv 0 mod 5 ). For example, ( a equiv 1 mod 5 ) and ( b equiv 3 mod 5 ). This shows 5 can be a factor. Higher power 25 can also be achieved (e.g., ( a = 1 ), ( b = 3 ) gives ( 7a + 6b = 25 )).4. Prime 7: ( 7a + 6b equiv 6b mod 7 ). Since ( b ) is not divisible by 7, 7 cannot be a factor.Maximum GCD:The maximum GCD is ( 5^2 = 25 ), achieved when ( 7a + 6b ) is divisible by 25. For example, ( a = 1 ) and ( b = 3 ) gives ( 7 cdot 1 + 6 cdot 3 = 25 ).Minimum Denominator:[frac{4200}{25} = 168]Thus, the minimum value for the denominator of the sum of the fractions is (boxed{168}).

🔑:1. Identify the fractions: Consider two irreducible fractions with denominators (600) and (700), respectively. Let's denote these fractions as (frac{a}{600}) and (frac{b}{700}), where (a) and (b) are integers that do not share common factors with (600) and (700) respectively (other than 1).2. Express the sum of these fractions: [ frac{a}{600} + frac{b}{700} ]3. Find a common denominator: The least common multiple (LCM) of (600) and (700) needs to be determined for the sum: - Prime factorization of (600): [ 600 = 2^3 times 3 times 5^2 ] - Prime factorization of (700): [ 700 = 2^2 times 5^2 times 7 ] The LCM takes the highest powers of all primes involved: [ text{LCM}(600, 700) = 2^3 times 3 times 5^2 times 7 = 4200 ]4. Convert each fraction to have the common denominator: [ frac{a}{600} = frac{a times 7}{600 times 7} = frac{7a}{4200} ] [ frac{b}{700} = frac{b times 6}{700 times 6} = frac{6b}{4200} ] Thus: [ frac{a}{600} + frac{b}{700} = frac{7a}{4200} + frac{6b}{4200} = frac{7a + 6b}{4200} ]5. Analyze the numerator for simplification: The expression: [ frac{7a+6b}{4200} = frac{7a+6b}{2^3 times 3 times 5^2 times 7} ] To simplify this fraction, let's examine the divisors of the numerator (7a + 6b): - Since (a) has no common factors with (600), (a) is coprime with (2^3, 3, 5^2). - Similarly, (b) has no common factors with (700), so (b) is coprime with (2^2, 5^2, 7).6. Select (a) and (b) to minimize the result: To identify (a) and (b) such that the fraction is simplified: - We want (7a + 6b) to be minimized. - Let's choose (a = 1) and (b = 3) (as suggestive values to check the minimal simplified numerator): [ 7a + 6b = 7 times 1 + 6 times 3 = 7 + 18 = 25 ] With this chosen (a) and (b): [ frac{7a + 6b}{4200} = frac{25}{4200} = frac{1}{168} ]7. Conclusion: The minimum denominator of the sum of the two fractions can be simplified to: [ boxed{168} ]Thus, the minimum possible denominator of the sum is ( boxed{168} ).

🧠:Okay, so I need to find the modulus of the complex number z that satisfies the equation (2017z - 25)/(z - 2017) = 3 + 4i. Hmm, let's start by recalling that the modulus of a complex number a + bi is sqrt(a² + b²). But first, I need to solve for z from the given equation. Let me write down the equation again to visualize it better:(2017z - 25)/(z - 2017) = 3 + 4iAlright, so to solve for z, I can cross-multiply to get rid of the denominator. Let me do that step by step. Multiply both sides by (z - 2017):2017z - 25 = (3 + 4i)(z - 2017)Now, I need to expand the right-hand side. Let me distribute (3 + 4i) over (z - 2017). Let's compute that:(3 + 4i)(z) - (3 + 4i)(2017)So, that becomes 3z + 4iz - 3*2017 - 4i*2017Let me compute the constants first. 3*2017 is 6051, and 4*2017 is 8068, so that term becomes -6051 - 8068i.Putting it all together, the right-hand side is 3z + 4iz - 6051 - 8068i.So, the equation now is:2017z - 25 = 3z + 4iz - 6051 - 8068iNow, let's get all terms involving z on the left side and constants on the right side. Subtract 3z and 4iz from both sides:2017z - 25 - 3z - 4iz = -6051 - 8068iSimplify the left side:(2017z - 3z - 4iz) - 25 = -6051 - 8068iFactor out z from the terms with z:z(2017 - 3 - 4i) - 25 = -6051 - 8068iCompute 2017 - 3, which is 2014, so:z(2014 - 4i) - 25 = -6051 - 8068iNow, add 25 to both sides to isolate the term with z:z(2014 - 4i) = -6051 - 8068i + 25Compute the constants on the right side:-6051 + 25 = -6026So, the right side becomes -6026 - 8068iTherefore, the equation is:z(2014 - 4i) = -6026 - 8068iNow, to solve for z, divide both sides by (2014 - 4i). So,z = (-6026 - 8068i)/(2014 - 4i)Hmm, complex division. To compute this, I can multiply numerator and denominator by the conjugate of the denominator. The conjugate of (2014 - 4i) is (2014 + 4i). Let's do that.Multiply numerator and denominator by (2014 + 4i):z = [(-6026 - 8068i)(2014 + 4i)] / [(2014 - 4i)(2014 + 4i)]First, compute the denominator. It's a difference of squares, so (2014)^2 - (4i)^2. Since i² = -1, this becomes 2014² - 16*(-1) = 2014² + 16.Compute 2014 squared. Hmm, 2014 is 2000 + 14. So, (2000 + 14)^2 = 2000² + 2*2000*14 + 14² = 4,000,000 + 56,000 + 196 = 4,056,196. Then add 16: 4,056,196 + 16 = 4,056,212. So denominator is 4,056,212.Now, the numerator: (-6026 - 8068i)(2014 + 4i). Let's expand this.Multiply each term:First, -6026*2014, then -6026*4i, then -8068i*2014, then -8068i*4i.Let me compute each term step by step.1. Compute -6026*2014:This seems like a big multiplication. Let me break it down.2014 * 6000 = 12,084,000But since it's 6026, which is 6000 + 26, so:2014*6000 = 12,084,0002014*26 = 52, 364 (since 2000*26=52,000 and 14*26=364, total 52,000 + 364 = 52,364)So total 12,084,000 + 52,364 = 12,136,364But since it's -6026*2014, that's -12,136,3642. Compute -6026*4i:-6026*4 = -24,104, so this term is -24,104i3. Compute -8068i*2014:-8068*2014i. Let's compute 8068*2014 first.Again, break it down:2014*8000 = 16,112,0002014*68 = 2014*(60 + 8) = 2014*60 = 120,840; 2014*8 = 16,112Total: 120,840 + 16,112 = 136,952So total 16,112,000 + 136,952 = 16,248,952But since it's -8068*2014 = -16,248,952, so this term is -16,248,952i4. Compute -8068i*4i:Multiply coefficients: -8068*4 = -32,272Multiply i*i: i² = -1, so this term is -32,272*(-1) = +32,272So the fourth term is +32,272Now, adding all four terms together:First term: -12,136,364Second term: -24,104iThird term: -16,248,952iFourth term: +32,272Combine real parts: -12,136,364 + 32,272 = (-12,136,364 + 32,272) = -12,104,092Combine imaginary parts: (-24,104i -16,248,952i) = (-24,104 -16,248,952)i = -16,273,056iSo numerator is -12,104,092 -16,273,056iTherefore, z = (-12,104,092 -16,273,056i) / 4,056,212Now, let's simplify this fraction by dividing numerator and denominator by 2 to see if we can reduce the numbers.Wait, 4,056,212 divided by 2 is 2,028,106.Similarly, numerator: -12,104,092 / 2 = -6,052,046-16,273,056 / 2 = -8,136,528So, z = (-6,052,046 -8,136,528i)/2,028,106Now, let's check if we can divide numerator and denominator by another common factor. Let's see if 2,028,106 divides into numerator coefficients.Check if 2,028,106 divides into 6,052,046:6,052,046 ÷ 2,028,106 ≈ 3 (since 2,028,106*3 = 6,084,318 which is more than 6,052,046). So, no, not exactly. Wait, maybe 2,028,106 is a factor here? Let me check:Wait, perhaps the denominator is 2,028,106 and the numerator terms are -6,052,046 and -8,136,528.Wait, 6,052,046 divided by 2,028,106: Let's compute 2,028,106 * 3 = 6,084,318, which is larger than 6,052,046. So 2,028,106 * 2.98 approximately. Not an integer. Hmm.Alternatively, maybe the numbers can be simplified by noticing that 2014 is 2*19*53 (if I factor 2014). Wait, 2014 divided by 2 is 1007. Then 1007 divided by 19 is 53. So yes, 2014 = 2*19*53. But I don't know if that helps here.Alternatively, maybe check if 2014 and 4 have a common factor with the numerator terms. Wait, 2014 is in the original denominator. But maybe there's a better approach here.Wait, maybe instead of expanding out the entire numerator, there's a smarter way to compute the modulus |z| without having to compute z explicitly. Because modulus has a property that |z| = |numerator| / |denominator|.Yes! That's a key insight. Since z = (numerator)/(denominator), then |z| = |numerator| / |denominator|.So, instead of computing the entire complex division, I can compute the modulus of the numerator and denominator separately and divide them.Let me recall that for complex numbers a and b, |a/b| = |a| / |b|. So yes, this is a much better approach. That would save me from dealing with all that messy multiplication.So, starting again, after cross-multiplying and getting to z = (-6026 - 8068i)/(2014 - 4i)Then |z| = | -6026 -8068i | / | 2014 -4i |Compute | -6026 -8068i | = sqrt( (-6026)^2 + (-8068)^2 )Similarly, | 2014 -4i | = sqrt(2014^2 + (-4)^2 )So let's compute these two moduli.First, compute the denominator modulus:sqrt(2014² + 4²)2014 squared we already computed earlier as 4,056,1964 squared is 16So denominator modulus is sqrt(4,056,196 + 16) = sqrt(4,056,212)Which is the same as the denominator we had before, 2014² + 16 = 4,056,212, so sqrt(4,056,212). Let's note that.Now, numerator modulus: sqrt( (-6026)^2 + (-8068)^2 )Compute 6026² and 8068².First, 6026². Let's compute that.6026 is 6000 + 26.So (6000 + 26)^2 = 6000² + 2*6000*26 + 26² = 36,000,000 + 312,000 + 676 = 36,312,676Similarly, 8068 is 8000 + 68.So (8000 + 68)^2 = 8000² + 2*8000*68 + 68² = 64,000,000 + 1,088,000 + 4,624 = 65,092,624Therefore, the numerator modulus is sqrt(36,312,676 + 65,092,624) = sqrt(101,405,300)Wait, let's check:36,312,676 + 65,092,624 = 101,405,300. Yes.So sqrt(101,405,300). Hmm, let's see if this simplifies.Let me check if 101,405,300 is a perfect square or if it factors into something.Alternatively, perhaps 101,405,300 = 10,070²?Wait, 10,000² = 100,000,00010,070² = (10,000 + 70)^2 = 10,000² + 2*10,000*70 + 70² = 100,000,000 + 1,400,000 + 4,900 = 101,404,900Ah, very close to 101,405,300. The difference is 101,405,300 - 101,404,900 = 400.So 10,070² = 101,404,900Then 101,404,900 + 400 = 101,405,300 = (10,070)^2 + (20)^2? Wait, no, because 10,070² + 20² = 101,404,900 + 400 = 101,405,300. Hmm, but that doesn't directly help. Alternatively, maybe it's 10,070.0... squared? Let me check.Compute 10,070.01², but probably not. Alternatively, perhaps 10,070.0² = 101,404,900, then 10,070.0² + 400 = 101,405,300. So sqrt(101,405,300) is sqrt(10,070² + 20²). Wait, but that would be the hypotenuse of a right triangle with sides 10,070 and 20, so sqrt(10,070² + 20²). But unless there's a better way, maybe just approximate it or factor.Alternatively, let's factor 101,405,300. Let's see:101,405,300 = 100 * 1,014,053Hmm, 1,014,053. Let's check if this is a square.sqrt(1,014,053) ≈ 1007, because 1007² = 1,014,049. So 1007² = 1,014,049. Then 1,014,053 - 1,014,049 = 4. So 1,014,053 = 1007² + 4 = 1007² + 2². Therefore, sqrt(1,014,053) = sqrt(1007² + 2²). That's again the hypotenuse of a right triangle.Therefore, sqrt(1,014,053) = sqrt(1007² + 2²). Not a whole number. So, 1,014,053 is not a perfect square, so 101,405,300 = 100 * 1,014,053 = 100*(1007² + 2²). Therefore, sqrt(101,405,300) = sqrt(100*(1007² + 2²)) = 10*sqrt(1007² + 2²). Hmm, which doesn't simplify further. So perhaps we can leave it as sqrt(101,405,300), but let's see if we can relate it to the denominator's modulus.Wait, the denominator's modulus was sqrt(4,056,212). Let me check if there's a relation between 101,405,300 and 4,056,212.Notice that 101,405,300 divided by 4,056,212 is approximately 25. Let me compute:4,056,212 * 25 = 101,405,300. Exactly! Wow, that's a key insight.So sqrt(101,405,300) = sqrt(25 * 4,056,212) = 5 * sqrt(4,056,212)Therefore, |numerator| = 5 * sqrt(4,056,212)And |denominator| = sqrt(4,056,212)Therefore, |z| = |numerator| / |denominator| = (5 * sqrt(4,056,212)) / sqrt(4,056,212) ) = 5So, the modulus of z is 5.Wait, that's a very clean answer. Let me verify this again because sometimes these modulus ratios can give such results. Let me recap:We had z = (-6026 -8068i)/(2014 -4i)Compute |z| = | -6026 -8068i | / | 2014 -4i |We computed | -6026 -8068i | = sqrt(6026² + 8068²) = sqrt(36,312,676 + 65,092,624) = sqrt(101,405,300)And |2014 -4i| = sqrt(2014² + 4²) = sqrt(4,056,196 + 16) = sqrt(4,056,212)Then noticed that 101,405,300 = 25 * 4,056,212, so sqrt(25 * 4,056,212) = 5*sqrt(4,056,212)Therefore, dividing gives 5*sqrt(4,056,212)/sqrt(4,056,212) = 5. So |z| = 5. That's brilliant. So despite the messy numbers, the modulus simplifies to 5. That seems correct.Alternatively, let me check the cross-multiplied equation again. Maybe there's another way to see that |z| = 5.Alternatively, think of the original equation as a Möbius transformation. The equation (2017z -25)/(z -2017) = 3 +4i. Perhaps there's a geometric interpretation here. But maybe that's more complicated.Alternatively, if we let w = z -2017, then express the equation in terms of w. But that might not necessarily help.Alternatively, consider that cross-multiplying gives 2017z -25 = (3 +4i)(z -2017). Then rearrange terms to get 2017z - (3 +4i)z = -25 + (3 +4i)*2017Factor z: z(2017 -3 -4i) = -25 + (3 +4i)*2017Which is the same as before, leading to z = [ -25 + (3 +4i)*2017 ] / (2014 -4i )But maybe if I compute the numerator modulus here, it's the same as before. Let's see:Numerator: -25 + (3 +4i)*2017Compute modulus:| -25 + 3*2017 + 4i*2017 | = | (3*2017 -25) + i*(4*2017) |Compute 3*2017 = 6051, so 6051 -25 = 60264*2017 = 8068So the numerator is 6026 + 8068i, but in our previous case, we had -6026 -8068i. But modulus is the same because |-a -bi| = |a + bi|. So modulus sqrt(6026² + 8068²) which is the same as before. So indeed, same result.Therefore, |z| is indeed 5. So the answer is 5.Final Answerboxed{5}

✅:Given the complex number ( z ) that satisfies the equation[frac{2017z - 25}{z - 2017} = 3 + 4i,]we need to determine the modulus ( |z| ).First, we cross-multiply to eliminate the denominator:[2017z - 25 = (3 + 4i)(z - 2017).]Expanding the right-hand side:[(3 + 4i)(z - 2017) = 3z + 4iz - 3 cdot 2017 - 4i cdot 2017.]Calculating the constants:[3 cdot 2017 = 6051 quad text{and} quad 4 cdot 2017 = 8068.]Thus, the equation becomes:[2017z - 25 = 3z + 4iz - 6051 - 8068i.]Rearranging terms to isolate ( z ):[2017z - 3z - 4iz = -6051 - 8068i + 25.]Simplifying:[z(2014 - 4i) = -6026 - 8068i.]Solving for ( z ):[z = frac{-6026 - 8068i}{2014 - 4i}.]To find the modulus ( |z| ), we use the property of moduli for complex numbers:[|z| = frac{|-6026 - 8068i|}{|2014 - 4i|}.]Calculating the moduli:[|-6026 - 8068i| = sqrt{(-6026)^2 + (-8068)^2} = sqrt{6026^2 + 8068^2},][|2014 - 4i| = sqrt{2014^2 + (-4)^2}.]Computing the squares:[6026^2 = 36312676, quad 8068^2 = 65092624, quad 2014^2 = 4056196, quad 4^2 = 16.]Summing the squares:[6026^2 + 8068^2 = 36312676 + 65092624 = 101405300,][2014^2 + 4^2 = 4056196 + 16 = 4056212.]Taking the square roots:[sqrt{101405300} = 5 sqrt{4056212},][sqrt{4056212}.]Thus, the modulus is:[|z| = frac{5 sqrt{4056212}}{sqrt{4056212}} = 5.]The final answer is:[boxed{5}]

🔑:1. Let ( w = 3 + 4mathrm{i} ), as given in the problem.2. The given equation is: [ frac{2017z - 25}{z - 2017} = 3 + 4mathrm{i} ] Substitute ( w = 3 + 4mathrm{i} ) into the equation: [ frac{2017z - 25}{z - 2017} = w ]3. Cross-multiply to clear the fraction: [ 2017z - 25 = w (z - 2017) ]4. Expand the right-hand side: [ 2017z - 25 = wz - 2017w ]5. Rearrange the terms to isolate ( z ): [ 2017z - wz = 2017w - 25 ]6. Factor out ( z ) on the left-hand side: [ z (2017 - w) = 2017w - 25 ]7. Solve for ( z ): [ z = frac{2017w - 25}{2017 - w} ]8. Calculate ( w bar{w} ) where (bar{w}) is the complex conjugate of ( w ): [ bar{w} = 3 - 4mathrm{i} ] [ w bar{w} = (3 + 4mathrm{i})(3 - 4mathrm{i}) = 3^2 - (4mathrm{i})^2 = 9 - 16(-1) = 9 + 16 = 25 ]9. Substitute ( w bar{w} ) back into the equation for ( z ): [ z = frac{25 - 2017w}{2017 - w} ]10. Simplify the expression: [ z = -w cdot frac{2017 - bar{w}}{2017 - w} ]11. Compute the magnitudes: [ |z| = |-w| cdot left|frac{2017 - bar{w}}{2017 - w}right| ]12. Since ( |-w| = |w| ) and ( |w| = sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5 ), then: [ |z| = 5 ] Conclusion:[boxed{5}]

🧠:Okay, let's try to solve this problem. So, we have an acute triangle with angles α, β, γ opposite sides a, b, c respectively. We're given that sin α = 3/5 and cos β = 5/13. We need to evaluate the expression (a² + b² - c²)/(ab). Hmm, that expression looks familiar. Wait, isn't that related to the Law of Cosines? Let me recall. The Law of Cosines states that c² = a² + b² - 2ab cos γ. So, if we rearrange that, we get a² + b² - c² = 2ab cos γ. Therefore, the expression (a² + b² - c²)/(ab) would simplify to 2 cos γ. So, the problem reduces to finding cos γ, since once we have that, we can just multiply by 2 to get the answer. Alright, so the key is to find cos γ. Since the triangle is acute, all angles are less than 90 degrees, so their cosines will be positive. Let's remember that in a triangle, the sum of the angles is π radians (or 180 degrees). So, α + β + γ = π. Therefore, γ = π - α - β. So, cos γ = cos(π - α - β). Using the cosine of a sum formula, cos(π - α - β) = -cos(α + β) because cos(π - θ) = -cos θ. Wait, actually, let me verify that. Wait, cos(π - θ) = -cos θ. But here, we have cos(π - (α + β)) which is equal to -cos(α + β). Therefore, cos γ = -cos(α + β). But since we need to compute cos γ, we can use the formula for cos(α + β): cos α cos β - sin α sin β. Therefore, cos γ = - [cos α cos β - sin α sin β] = -cos α cos β + sin α sin β. So, cos γ = sin α sin β - cos α cos β. Therefore, 2 cos γ would be 2(sin α sin β - cos α cos β). But wait, let's double-check that. Wait, if γ = π - α - β, then cos γ = cos(π - α - β) = -cos(α + β). Then, using the cosine addition formula: cos(α + β) = cos α cos β - sin α sin β. Therefore, cos γ = - (cos α cos β - sin α sin β) = -cos α cos β + sin α sin β. So, yes, that part is correct. Therefore, to find cos γ, we need sin α, cos α, sin β, and cos β. Wait, we are given sin α = 3/5 and cos β = 5/13. So, we need to find cos α and sin β. Since α is an acute angle (because the triangle is acute), cos α will be positive. Similarly, β is acute, so sin β will be positive. So, let's compute cos α first. Since sin α = 3/5, we can use the identity sin² α + cos² α = 1. Therefore, cos² α = 1 - (3/5)² = 1 - 9/25 = 16/25. Therefore, cos α = 4/5. Similarly, for β, we have cos β = 5/13. So, sin β can be found using sin² β + cos² β = 1. Therefore, sin² β = 1 - (5/13)² = 1 - 25/169 = 144/169. So, sin β = 12/13. Now, plug these values into the expression for cos γ. cos γ = sin α sin β - cos α cos β= (3/5)(12/13) - (4/5)(5/13)= (36/65) - (20/65)= (36 - 20)/65= 16/65Therefore, 2 cos γ = 2*(16/65) = 32/65. So, the value of (a² + b² - c²)/(ab) is 32/65. Wait, but let me confirm if I did everything correctly. Let's check step by step. First, using Law of Cosines: c² = a² + b² - 2ab cos γ, so rearranged gives a² + b² - c² = 2ab cos γ. Therefore, (a² + b² - c²)/(ab) = 2 cos γ. That's correct. Then, since γ = π - α - β, so cos γ = -cos(α + β). Then, using the formula for cos(α + β) = cos α cos β - sin α sin β. Therefore, cos γ = -cos(α + β) = - (cos α cos β - sin α sin β) = sin α sin β - cos α cos β. That's correct. Then, computing cos α: sin α = 3/5, so cos α = sqrt(1 - (9/25)) = sqrt(16/25) = 4/5. Correct. For sin β: cos β = 5/13, so sin β = sqrt(1 - 25/169) = sqrt(144/169) = 12/13. Correct. Then plug into cos γ: (3/5)(12/13) - (4/5)(5/13) = (36/65 - 20/65) = 16/65. So, 2 cos γ = 32/65. Yes, that seems correct. But let's make sure that there isn't a mistake in the angle relations. Since in the Law of Cosines, the angle opposite side c is γ. So, c² = a² + b² - 2ab cos γ. So, the expression a² + b² - c² is equal to 2ab cos γ, so dividing by ab gives 2 cos γ. So, everything checks out. Alternatively, another way to approach this problem might be using the Law of Sines. Let's see if we can cross-verify. Law of Sines states that a/sin α = b/sin β = c/sin γ = 2R, where R is the radius of the circumscribed circle. Given that sin α = 3/5 and sin β = 12/13 (we found this earlier), and cos α = 4/5, cos β = 5/13. So, suppose we denote the common ratio as 2R. Then, a = 2R sin α = 2R*(3/5), b = 2R*(12/13), and c = 2R sin γ. Therefore, a = (6R)/5, b = (24R)/13, c = 2R sin γ. But maybe this approach complicates things, unless we can find sin γ. However, since we already found cos γ = 16/65, then sin γ can be found as sqrt(1 - (16/65)^2) = sqrt(1 - 256/4225) = sqrt(3969/4225) = 63/65. Therefore, c = 2R*(63/65). But perhaps we can compute a, b, c in terms of R and then compute (a² + b² - c²)/(ab). Let's try that. Compute a² + b² - c²:a² = (6R/5)^2 = 36R²/25b² = (24R/13)^2 = 576R²/169c² = (2R*63/65)^2 = (126R/65)^2 = 15876R²/4225Therefore, a² + b² - c² = (36R²/25 + 576R²/169 - 15876R²/4225)Let's compute each term with denominator 4225:36R²/25 = (36*169 R²)/4225 = (6084 R²)/4225576R²/169 = (576*25 R²)/4225 = (14400 R²)/422515876R²/4225 is already in that denominator.So, a² + b² - c² = (6084 + 14400 - 15876) R² / 4225Compute numerator:6084 + 14400 = 2048420484 - 15876 = 4608Therefore, a² + b² - c² = 4608 R² / 4225Now, compute ab:a = 6R/5, b = 24R/13ab = (6R/5)(24R/13) = (144 R²)/65Therefore, (a² + b² - c²)/(ab) = (4608 R² /4225) / (144 R² /65) = (4608 /4225) / (144 /65) * (R²/R²) = (4608 /4225) * (65 /144)Simplify:4608 ÷ 144 = 3265 ÷ 4225 = 65 ÷ (65*65) = 1/65Wait, 4225 = 65*65, right? Because 65*60=3900, 65*5=325, so 3900+325=4225. Yes.So, 4608 /4225 = 4608 / (65*65)144 = 144So, (4608 /4225) * (65/144) = (4608 *65)/(4225*144) But 4225 = 65*65, so:= (4608 *65)/(65*65*144) = 4608/(65*144)4608 ÷ 144 = 32Therefore, 32/65.So, (a² + b² - c²)/(ab) = 32/65. Which matches our previous result. Therefore, this confirms that the answer is indeed 32/65. So, after going through both methods—using the Law of Cosines directly and then verifying with the Law of Sines—we arrive at the same answer. Therefore, I'm confident that 32/65 is the correct result.Final Answerboxed{dfrac{32}{65}}

✅:Given an acute triangle with angles (alpha, beta, gamma) opposite sides (a, b, c) respectively, we know (sin alpha = frac{3}{5}) and (cos beta = frac{5}{13}). We need to evaluate (frac{a^{2} + b^{2} - c^{2}}{ab}).First, we use the Law of Cosines, which states (c^2 = a^2 + b^2 - 2ab cos gamma). Rearranging this, we get (a^2 + b^2 - c^2 = 2ab cos gamma). Therefore, the expression simplifies to (2 cos gamma).Next, we find (cos gamma). Since (gamma = pi - alpha - beta), we use the identity (cos gamma = cos(pi - alpha - beta) = -cos(alpha + beta)). Using the cosine addition formula, (cos(alpha + beta) = cos alpha cos beta - sin alpha sin beta), we get:[cos gamma = -(cos alpha cos beta - sin alpha sin beta) = sin alpha sin beta - cos alpha cos beta]We calculate (cos alpha) and (sin beta) using the given values:[cos alpha = sqrt{1 - left(frac{3}{5}right)^2} = sqrt{frac{16}{25}} = frac{4}{5}][sin beta = sqrt{1 - left(frac{5}{13}right)^2} = sqrt{frac{144}{169}} = frac{12}{13}]Substituting these values into the expression for (cos gamma):[cos gamma = left(frac{3}{5} cdot frac{12}{13}right) - left(frac{4}{5} cdot frac{5}{13}right) = frac{36}{65} - frac{20}{65} = frac{16}{65}]Thus, (2 cos gamma = 2 cdot frac{16}{65} = frac{32}{65}).The final answer is:[boxed{dfrac{32}{65}}]

🔑:Given:- alpha is an angle in an acute triangle with sinalpha = frac{3}{5}.- beta is another angle in the same triangle with cosbeta = frac{5}{13}.Our goal is to find the value of frac{a^2 + b^2 - c^2}{ab}, where a, b, c are the sides opposite angles alpha, beta, gamma respectively.1. Calculate cosalpha and sinbeta: From sinalpha = frac{3}{5}, we use the Pythagorean identity: [ cosalpha = sqrt{1 - sin^2 alpha} = sqrt{1 - left(frac{3}{5}right)^2} = sqrt{1 - frac{9}{25}} = sqrt{frac{16}{25}} = frac{4}{5} ] Since beta is an angle in the same triangle and cosbeta = frac{5}{13}, we use the Pythagorean identity: [ sinbeta = sqrt{1 - cos^2 beta} = sqrt{1 - left(frac{5}{13}right)^2} = sqrt{1 - frac{25}{169}} = sqrt{frac{144}{169}} = frac{12}{13} ]2. Use the Law of Cosines in the triangle: The Law of Cosines states: [ c^2 = a^2 + b^2 - 2ab cosgamma ] We look to find cosgamma. Since alpha + beta + gamma = pi (the sum of angles in a triangle): [ gamma = pi - alpha - beta ] Using the sine rule for sin(pi - text{(angle)}) = sin(text{(angle)}), we can find cosgamma using addition formulas for angles: [ cosgamma = cos(pi - alpha - beta) = -cos(alpha + beta) ] Now using the trigonometric identity for sum of angles: [ cos(alpha + beta) = cosalpha cosbeta - sinalpha sinbeta ] Substitute the values: [ cos(alpha + beta) = left(frac{4}{5}right)left(frac{5}{13}right) - left(frac{3}{5}right)left(frac{12}{13}right) ] [ = frac{4 cdot 5}{65} - frac{3 cdot 12}{65} ] [ = frac{20}{65} - frac{36}{65} ] [ = frac{20 - 36}{65} ] [ = -frac{16}{65} ] Thus, [ cos(pi - alpha - beta) = -left(-frac{16}{65}right) = frac{16}{65} ]3. Calculate frac{a^2 + b^2 - c^2}{ab}: Using the Law of Cosines formula adjusted: [ c^2 = a^2 + b^2 - 2ab left(frac{16}{65}right) ] We rearrange to find the desired expression: [ a^2 + b^2 - c^2 = 2ab cosgamma = 2ab left(frac{16}{65}right) ] [ = 2ab left(frac{16}{65}right) ] [ = frac{32ab}{65} ] Therefore, the ratio: [ frac{a^2 + b^2 - c^2}{ab} = frac{frac{32ab}{65}}{ab} ] [ = frac{32}{65} ]Conclusion. [boxed{frac{32}{65}}]