💡Okay, so I have this multiple-choice question here with five statements, and I need to figure out which ones are correct. The options are labeled ① through ⑤, and then the choices A through D each have different combinations of these statements. Let me take it step by step.First, let me look at each statement one by one.**Statement ①: Inductive reasoning is reasoning from specific to general.**Hmm, I remember that inductive reasoning involves making broad generalizations from specific observations. So, if you observe specific instances and then draw a general conclusion, that's inductive reasoning. For example, if I notice that the sun rises every morning, I might conclude that the sun rises every day. That seems like reasoning from specific observations to a general principle. So, I think statement ① is correct.**Statement ②: Deductive reasoning is reasoning from general to specific.**Deductive reasoning, on the other hand, starts with a general statement or hypothesis and examines the possibilities to reach a specific, logical conclusion. For instance, if all men are mortal and Socrates is a man, then Socrates is mortal. That's moving from a general principle to a specific case. So, statement ② also seems correct.**Statement ③: Analogical reasoning is reasoning from specific to general.**Analogical reasoning involves comparing two things and drawing conclusions based on their similarities. It's about transferring knowledge from one domain to another. For example, if a new kind of fruit tastes similar to an apple, you might reason that it's also good in pies. But is this reasoning from specific to general? Or is it more about comparing specific cases?Wait, analogical reasoning is actually about comparing specific instances to draw conclusions about another specific instance, not necessarily moving from specific to general. So, maybe statement ③ is incorrect. It doesn't go from specific to general; it's more about specific to specific by analogy.**Statement ④: The analytical method is an indirect proof method.**The analytical method, as I recall, is a direct method of proof where you start with what you know and use logical steps to reach the conclusion. Indirect proof methods, like proof by contradiction, assume the opposite and show it leads to a contradiction. So, if the analytical method is direct, then statement ④ is incorrect.**Statement ⑤: If ( z in mathbb{C} ), and ( |z + 2 - 2i| = 1 ), then the minimum value of ( |z - 2 - 2i| ) is 3.**Alright, this is a complex numbers problem. Let me visualize it. The equation ( |z + 2 - 2i| = 1 ) represents a circle in the complex plane with center at ( (-2, 2) ) and radius 1. The expression ( |z - 2 - 2i| ) represents the distance from the point ( z ) to the point ( (2, 2) ).So, we need to find the minimum distance from any point on the circle ( |z + 2 - 2i| = 1 ) to the point ( (2, 2) ). To find this, I can calculate the distance between the center of the circle ( (-2, 2) ) and the point ( (2, 2) ), and then subtract the radius of the circle.The distance between ( (-2, 2) ) and ( (2, 2) ) is ( |2 - (-2)| = 4 ) units along the real axis. Since the radius is 1, the minimum distance from the circle to the point ( (2, 2) ) is ( 4 - 1 = 3 ). So, statement ⑤ is correct.Now, let me recap:- ①: Correct- ②: Correct- ③: Incorrect- ④: Incorrect- ⑤: CorrectSo, the correct statements are ①, ②, and ⑤. Looking back at the options:A: ①②③④ B: ②③④ C: ①②④⑤ D: ①②⑤Therefore, the correct choice is D: ①②⑤.**Final Answer**boxed{D}

💡First, I need to determine the total number of plates available. There are 8 red plates and 6 blue plates, making a total of 14 plates.Next, I'll calculate the total number of ways to choose any two plates from these 14 plates using the combination formula ( binom{14}{2} ). This gives me ( frac{14 times 13}{2} = 91 ) possible pairs.Then, I'll find the number of ways to choose two red plates from the 8 available. Using the combination formula ( binom{8}{2} ), I get ( frac{8 times 7}{2} = 28 ) pairs of red plates.Similarly, I'll calculate the number of ways to choose two blue plates from the 6 available using ( binom{6}{2} ), which results in ( frac{6 times 5}{2} = 15 ) pairs of blue plates.Finally, I'll add the successful outcomes of choosing two red plates and two blue plates to get a total of 43 successful pairs. The probability of selecting two plates of the same color is then ( frac{43}{91} ).

💡Okay, so I need to find the angle between two vectors, vector a and vector b. Vector a is given as (√3, 1) and vector b is (√3, -1). Hmm, I remember that to find the angle between two vectors, I can use the dot product formula. Let me recall the formula... I think it's something like the dot product of a and b equals the product of their magnitudes times the cosine of the angle between them. So, in symbols, that would be:a · b = |a| |b| cosθWhere θ is the angle between them. So, if I can find the dot product and the magnitudes of both vectors, I can solve for θ.First, let's find the dot product of a and b. The dot product is calculated by multiplying the corresponding components and then adding them together. So, for vector a = (√3, 1) and vector b = (√3, -1), the dot product would be:a · b = (√3)(√3) + (1)(-1)Let me compute that. √3 times √3 is 3, and 1 times -1 is -1. So, adding those together: 3 + (-1) = 2. So, the dot product is 2.Next, I need to find the magnitudes of both vectors a and b. The magnitude of a vector is found using the square root of the sum of the squares of its components. So, for vector a:|a| = √[(√3)^2 + (1)^2] = √[3 + 1] = √4 = 2Similarly, for vector b:|b| = √[(√3)^2 + (-1)^2] = √[3 + 1] = √4 = 2So, both vectors have a magnitude of 2.Now, plugging these values into the dot product formula:2 = (2)(2) cosθSimplifying the right side:2 = 4 cosθTo solve for cosθ, divide both sides by 4:cosθ = 2 / 4 = 1/2Now, I need to find the angle θ whose cosine is 1/2. I remember that cos(π/3) = 1/2. So, θ must be π/3 radians.Wait, let me just double-check. Cosine of 60 degrees is 0.5, which is 1/2, and 60 degrees is π/3 radians. So, that seems correct.Is there any chance I made a mistake in the calculations? Let me go through each step again.Dot product: (√3)(√3) + (1)(-1) = 3 -1 = 2. That's correct.Magnitudes: Both vectors have components squared as 3 and 1, so √(3+1)=2. That's correct.Then, 2 = 4 cosθ => cosθ = 1/2. Yes, that's right.And θ is π/3. Yeah, that seems solid.I don't think I made any errors here. So, the angle between the two vectors is π/3 radians.**Final Answer**The angle between the vectors is boxed{dfrac{pi}{3}}.

💡Okay, so I have this problem here about sets A and B, and I need to find their intersection, A ∩ B. Let me try to understand what each set represents first.Set A is defined as {x | x ≥ -2}. That means A includes all real numbers x that are greater than or equal to -2. So, if I imagine a number line, A starts at -2 and goes all the way to the right, including -2 itself.Set B is defined as {x | x < 1}. This means B includes all real numbers x that are less than 1. On the number line, B would start from the far left and go up to, but not including, 1.Now, to find the intersection of A and B, which is A ∩ B, I need to find all the numbers that are in both sets A and B. In other words, I'm looking for numbers that are both greater than or equal to -2 and less than 1.Let me visualize this on the number line. Set A covers from -2 to infinity, and set B covers from negative infinity up to 1. The overlap between these two sets would be from -2 up to, but not including, 1. So, the intersection should be all numbers x such that -2 ≤ x < 1.Looking at the options provided:A: {x | -2 ≤ x < -1} – Hmm, this seems too restrictive because it only goes up to -1, not 1.B: {-2, 1} – This is just two specific numbers, but the intersection should be a range of numbers, not just two points.C: {x | -2 < x < 1} – This is close, but it doesn't include -2 since it's strictly greater than -2. However, set A includes -2, so the intersection should also include -2.D: {x | -2 ≤ x ≤ 1} – This includes all numbers from -2 to 1, but set B doesn't include 1, so this can't be right either.Wait, none of the options seem to perfectly match what I thought. Let me double-check my understanding. Set A includes -2, and set B goes up to but doesn't include 1. So the intersection should include -2 and all numbers up to, but not including, 1. That would be {x | -2 ≤ x < 1}.Looking back at the options, option A is {x | -2 ≤ x < -1}, which is incorrect because it stops at -1. Option C is {x | -2 < x < 1}, which excludes -2, but since A includes -2, the intersection should include it. Option D includes 1, which B doesn't, so that's wrong too.Hmm, maybe I made a mistake. Let me think again. The intersection should include all x that are in both A and B. So x has to be ≥ -2 and < 1. That means x can be -2, but can't be 1. So the correct interval should be [-2, 1), which is written as {x | -2 ≤ x < 1}.But none of the options exactly say that. Option A is too narrow, stopping at -1. Option C is almost right but doesn't include -2. Option D includes 1, which isn't allowed. Maybe I need to reconsider.Wait, perhaps I misread the options. Let me check again:A: {x | -2 ≤ x < -1} – No, that's only up to -1.B: {-2, 1} – Just two points, not the interval.C: {x | -2 < x < 1} – Excludes -2.D: {x | -2 ≤ x ≤ 1} – Includes 1, which isn't in B.So none of the options perfectly match. But if I have to choose the closest one, option C is almost correct except it doesn't include -2. However, since set A does include -2, the intersection should include it. Maybe the options are slightly off, or perhaps I'm misunderstanding something.Alternatively, maybe the question has a typo, or the options are presented differently. But based on the given options, the best fit would be option C, even though it doesn't include -2. Or perhaps I'm overcomplicating it.Wait, let me think differently. Maybe the intersection is from -2 to 1, including -2 but not 1. So, in interval notation, that's [-2, 1). Looking at the options, none of them exactly match this. Option A is [-2, -1), which is too small. Option C is (-2, 1), which excludes -2. Option D is [-2, 1], which includes 1.Hmm, this is confusing. Maybe the correct answer isn't listed, but I have to choose the closest one. If I have to pick between C and D, C is better because it excludes 1, which is correct, but it also excludes -2, which isn't right. D includes 1, which is wrong. So maybe the answer is none of the above, but that's not an option.Wait, perhaps I made a mistake in interpreting the sets. Let me re-examine the definitions:A = {x | x ≥ -2} – So x can be -2, -1, 0, 0.5, etc., up to infinity.B = {x | x < 1} – So x can be any number less than 1, like 0, 0.5, but not 1.So the intersection should include all x that are ≥ -2 and < 1. That is, from -2 up to but not including 1.Looking back at the options:A: {x | -2 ≤ x < -1} – This is only from -2 to -1, which is too limited.B: {-2, 1} – Just two points, not the interval.C: {x | -2 < x < 1} – This is the interval from -2 to 1, excluding both endpoints.D: {x | -2 ≤ x ≤ 1} – This includes both -2 and 1.So, the correct interval should include -2 but exclude 1. None of the options do that exactly. Option C excludes -2, which is incorrect because A includes -2. Option D includes 1, which is incorrect because B doesn't include 1.Wait, maybe I'm overcomplicating. Perhaps the answer is option C, even though it excludes -2, because it's the closest to the correct interval. Or maybe the question expects the answer to be option C, assuming that -2 is not included, but that contradicts set A.Alternatively, perhaps the answer is option D, but that includes 1, which is not in B. So that can't be right.I'm stuck. Maybe I should look for another approach. Let's consider specific numbers:- Is -2 in both A and B? Yes, because A includes -2 and B doesn't exclude it (since B is x < 1, and -2 < 1). So -2 is in the intersection.- Is 0 in both A and B? Yes, because 0 ≥ -2 and 0 < 1.- Is 0.5 in both? Yes.- Is 1 in the intersection? No, because B doesn't include 1.- Is -3 in the intersection? No, because A doesn't include -3.So the intersection includes all numbers from -2 up to, but not including, 1. That is, [-2, 1).Looking at the options again:A: [-2, -1) – Too small.B: {-2, 1} – Not an interval.C: (-2, 1) – Excludes -2.D: [-2, 1] – Includes 1.None of these exactly match [-2, 1). But if I have to choose the closest, option C is the interval from -2 to 1, excluding both, but since -2 is included in A, the correct intersection should include -2. So maybe the answer is not listed correctly.Alternatively, perhaps the question expects the answer to be option C, assuming that -2 is not included, but that's incorrect because A includes -2.Wait, maybe I'm misinterpreting the notation. Let me check:Set A is {x | x ≥ -2}, so x can be -2, -1.5, -1, etc.Set B is {x | x < 1}, so x can be -2, -1, 0, 0.5, etc.So the intersection is all x such that x ≥ -2 and x < 1, which is [-2, 1).Looking at the options, none of them exactly match this. Option C is (-2, 1), which excludes -2, but since -2 is in both A and B, it should be included. Option D includes 1, which is not in B.Therefore, the correct answer isn't listed, but if I have to choose the closest, option C is better than D because it excludes 1, which is correct, but it also incorrectly excludes -2. Alternatively, maybe the answer is option A, but that's only up to -1, which is too limited.Wait, perhaps the answer is option C, and the exclusion of -2 is a mistake in the options. Or maybe I'm misunderstanding the original sets.Alternatively, maybe the answer is option D, but that includes 1, which is incorrect.I'm confused. Maybe I should go with option C, even though it's not perfect, because it's the only one that excludes 1, which is correct, even though it also excludes -2, which is not.Alternatively, perhaps the answer is none of the above, but that's not an option.Wait, let me think again. The intersection should include -2 because A includes it, and B doesn't exclude it. So the correct interval is [-2, 1). None of the options exactly match this. Option A is too small, option C excludes -2, option D includes 1.Maybe the answer is option C, assuming that the exclusion of -2 is a typo, or perhaps the question expects the answer to be option C.Alternatively, maybe I'm overcomplicating, and the answer is option C.Wait, no, I think I need to stick with the correct interval, which is [-2, 1). Since none of the options exactly match, but option C is the closest, excluding 1, which is correct, even though it excludes -2, which is not. But perhaps the answer is option C.Alternatively, maybe the answer is option A, but that's only up to -1, which is incorrect.I'm stuck. Maybe I should look for another way to approach this.Wait, perhaps the answer is option C, because it's the only one that represents an interval between -2 and 1, excluding 1, which is correct. Even though it excludes -2, which is in A, maybe the answer is option C.Alternatively, perhaps the answer is option D, but that includes 1, which is incorrect.I think I'll go with option C, even though it's not perfect, because it's the closest to the correct interval.

💡Okay, so I have this problem where I need to figure out how many ways there are to put 8 indistinguishable balls into 3 distinguishable boxes. The catch is that Box 1 must have at least 2 balls. Alright, let's break this down step by step.First, I remember that when dealing with indistinguishable objects and distinguishable boxes, the problem is a classic stars and bars problem. The formula for the number of ways to distribute n indistinguishable objects into k distinguishable boxes is C(n + k - 1, k - 1), where C is the combination function. But wait, that's when there are no restrictions. In this case, there's a restriction: Box 1 must have at least 2 balls. So, I need to adjust for that.I think the way to handle this is to first satisfy the condition for Box 1 and then distribute the remaining balls without any restrictions. So, if Box 1 needs at least 2 balls, I'll start by placing 2 balls in Box 1. That leaves me with 8 - 2 = 6 balls to distribute among the 3 boxes. Now, these 6 balls can go into any of the boxes, including Box 1, without any restrictions.So now, the problem reduces to distributing 6 indistinguishable balls into 3 distinguishable boxes. Using the stars and bars formula, the number of ways to do this is C(6 + 3 - 1, 3 - 1) = C(8, 2). Let me calculate that: C(8, 2) is 28. So, does that mean there are 28 ways?Wait, hold on. Let me make sure I'm not making a mistake here. So, the initial step was to place 2 balls in Box 1, which is correct because Box 1 must have at least 2 balls. Then, distributing the remaining 6 balls among the 3 boxes without any restrictions. That seems right.But just to double-check, maybe I can think of it differently. Suppose I don't place any balls in Box 1 first and then subtract the cases where Box 1 has fewer than 2 balls. So, the total number of ways without any restrictions is C(8 + 3 - 1, 3 - 1) = C(10, 2) = 45. Now, the number of ways where Box 1 has fewer than 2 balls is the number of ways where Box 1 has 0 or 1 ball.If Box 1 has 0 balls, then we're distributing all 8 balls into the remaining 2 boxes, which is C(8 + 2 - 1, 2 - 1) = C(9, 1) = 9.If Box 1 has 1 ball, then we're distributing the remaining 7 balls into the 3 boxes, which is C(7 + 3 - 1, 3 - 1) = C(9, 2) = 36.Wait, that doesn't seem right because 9 + 36 = 45, which is the total number of unrestricted distributions. But I'm supposed to subtract the cases where Box 1 has fewer than 2 balls, which would be 9 (for 0 balls) + 36 (for 1 ball) = 45. But that would mean the number of valid distributions is 45 - 45 = 0, which can't be right.Hmm, I must have made a mistake here. Let me re-examine this approach. When Box 1 has 0 balls, the number of ways is indeed C(8 + 2 - 1, 2 - 1) = 9. When Box 1 has 1 ball, the number of ways is C(7 + 3 - 1, 3 - 1) = C(9, 2) = 36. So, total invalid distributions are 9 + 36 = 45, which is the same as the total number of distributions without restrictions. That would imply that all distributions have Box 1 with fewer than 2 balls, which contradicts the initial problem statement that Box 1 must have at least 2 balls.This suggests that my second approach is flawed. Maybe I'm misapplying the formula when subtracting the invalid cases. Let me think again.Alternatively, perhaps the first approach was correct. By placing 2 balls in Box 1 first, we ensure that Box 1 has at least 2 balls, and then distributing the remaining 6 balls without restrictions. That gives us C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28 ways. This seems more plausible.Wait, but in the second approach, when I tried to subtract the invalid cases, I ended up with 0, which doesn't make sense. Maybe I misapplied the subtraction. Let me try again.Total number of distributions without restrictions: C(10, 2) = 45.Number of distributions where Box 1 has fewer than 2 balls: Number of distributions where Box 1 has 0 balls plus number where Box 1 has 1 ball.Number of distributions where Box 1 has 0 balls: C(8 + 2 - 1, 2 - 1) = C(9, 1) = 9.Number of distributions where Box 1 has 1 ball: C(7 + 3 - 1, 3 - 1) = C(9, 2) = 36.So, total invalid distributions: 9 + 36 = 45.Wait, that can't be right because 45 is the total number of distributions. That would mean there are no valid distributions where Box 1 has at least 2 balls, which contradicts the problem statement.This indicates that my second approach is incorrect. Maybe I'm miscounting the invalid distributions. Let me think differently.Perhaps when Box 1 has 1 ball, the remaining 7 balls are distributed among all 3 boxes, including Box 1. But since we've already fixed 1 ball in Box 1, the remaining 7 are distributed among all 3 boxes. So, the number of ways should be C(7 + 3 - 1, 3 - 1) = C(9, 2) = 36. That still seems correct.But then, adding the 9 ways where Box 1 has 0 balls, we get 45 invalid distributions, which is the same as the total. This suggests that all distributions have Box 1 with fewer than 2 balls, which can't be true because we can certainly have distributions where Box 1 has 2 or more balls.Therefore, my second approach must be flawed. Maybe I should stick with the first approach where I place 2 balls in Box 1 first and then distribute the remaining 6 balls, giving C(8, 2) = 28 ways.Alternatively, perhaps I'm misunderstanding the problem. Maybe the boxes are distinguishable, so the order matters, but the balls are indistinct. So, the number of ways should indeed be 28.Wait, let me try another way. Let's list out the possible distributions where Box 1 has at least 2 balls.Box 1 can have 2, 3, 4, 5, 6, 7, or 8 balls.For each case, we'll distribute the remaining balls among Box 2 and Box 3.- If Box 1 has 2 balls, remaining 6 balls: number of ways = C(6 + 2 - 1, 2 - 1) = C(7, 1) = 7.- If Box 1 has 3 balls, remaining 5 balls: C(5 + 2 - 1, 2 - 1) = C(6, 1) = 6.- If Box 1 has 4 balls, remaining 4 balls: C(4 + 2 - 1, 2 - 1) = C(5, 1) = 5.- If Box 1 has 5 balls, remaining 3 balls: C(3 + 2 - 1, 2 - 1) = C(4, 1) = 4.- If Box 1 has 6 balls, remaining 2 balls: C(2 + 2 - 1, 2 - 1) = C(3, 1) = 3.- If Box 1 has 7 balls, remaining 1 ball: C(1 + 2 - 1, 2 - 1) = C(2, 1) = 2.- If Box 1 has 8 balls, remaining 0 balls: C(0 + 2 - 1, 2 - 1) = C(1, 1) = 1.Now, adding these up: 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28.Okay, so this method also gives 28 ways, which matches the first approach. So, it seems that the correct answer is 28.Wait, but in my initial second approach, I got 0, which was clearly wrong. So, I must have made a mistake there. Maybe I confused the formula or misapplied it. It's clear from the first approach and the enumeration that the answer is 28.Therefore, the number of ways to distribute 8 indistinguishable balls into 3 distinguishable boxes with Box 1 having at least 2 balls is 28.

💡Okay, so I have this expression to simplify: it's a division of two fractions, each of which is a quadratic divided by another quadratic. Let me write it down again to make sure I have it right.The expression is:[frac{x^2 - 4x + 3}{x^2 - 6x + 8} div frac{x^2 - 6x + 9}{x^2 - 8x + 15}]Alright, I remember that dividing by a fraction is the same as multiplying by its reciprocal. So, I can rewrite this as:[frac{x^2 - 4x + 3}{x^2 - 6x + 8} times frac{x^2 - 8x + 15}{x^2 - 6x + 9}]Now, I need to factor each of these quadratics. Factoring quadratics is something I should be able to do. Let's take them one by one.Starting with the numerator of the first fraction: (x^2 - 4x + 3). I need two numbers that multiply to 3 and add up to -4. Hmm, -1 and -3. So, that factors to:[(x - 1)(x - 3)]Next, the denominator of the first fraction: (x^2 - 6x + 8). Looking for two numbers that multiply to 8 and add to -6. That would be -2 and -4. So, it factors to:[(x - 2)(x - 4)]Moving on to the numerator of the second fraction: (x^2 - 8x + 15). I need two numbers that multiply to 15 and add to -8. That would be -3 and -5. So, factoring gives:[(x - 3)(x - 5)]Lastly, the denominator of the second fraction: (x^2 - 6x + 9). This looks familiar; it's a perfect square. The square of (x - 3) is (x^2 - 6x + 9). So, it factors to:[(x - 3)^2]Now, substituting all these factored forms back into the expression, we have:[frac{(x - 1)(x - 3)}{(x - 2)(x - 4)} times frac{(x - 3)(x - 5)}{(x - 3)^2}]Before I proceed, I should note that (x) cannot be 2, 3, 4, or 5 because those values would make the denominator zero, which is undefined. So, I'll keep that in mind for the restrictions on (x).Now, let's multiply the numerators together and the denominators together:Numerator: ((x - 1)(x - 3)(x - 3)(x - 5))Denominator: ((x - 2)(x - 4)(x - 3)^2)Wait, actually, I think I made a mistake there. Let me correct that. When multiplying the numerators, it's ((x - 1)(x - 3)) times ((x - 3)(x - 5)), which is ((x - 1)(x - 3)^2(x - 5)). Similarly, the denominator is ((x - 2)(x - 4)(x - 3)^2).So, the expression becomes:[frac{(x - 1)(x - 3)^2(x - 5)}{(x - 2)(x - 4)(x - 3)^2}]Now, I can cancel out the common factors in the numerator and the denominator. I see that ((x - 3)^2) appears in both, so they cancel out. That leaves me with:[frac{(x - 1)(x - 5)}{(x - 2)(x - 4)}]Wait a second, looking back at the answer choices, option C is (frac{(x - 1)(x - 5)}{(x - 4)(x - 2)}), which is the same as what I have here. But hold on, I thought I had an extra ((x - 3)) in the denominator, but it seems like it canceled out.Let me double-check my factoring and cancellation steps to make sure I didn't make a mistake. First, factoring:- (x^2 - 4x + 3 = (x - 1)(x - 3)) – correct.- (x^2 - 6x + 8 = (x - 2)(x - 4)) – correct.- (x^2 - 8x + 15 = (x - 3)(x - 5)) – correct.- (x^2 - 6x + 9 = (x - 3)^2) – correct.Then, rewriting the expression:[frac{(x - 1)(x - 3)}{(x - 2)(x - 4)} times frac{(x - 3)(x - 5)}{(x - 3)^2}]Multiplying numerators and denominators:Numerator: ((x - 1)(x - 3)(x - 3)(x - 5))Denominator: ((x - 2)(x - 4)(x - 3)^2)Simplifying by canceling ((x - 3)^2):[frac{(x - 1)(x - 5)}{(x - 2)(x - 4)}]Yes, that seems correct. So, the simplified expression is (frac{(x - 1)(x - 5)}{(x - 2)(x - 4)}), which matches option C.But wait, looking back at the original problem, the division was of the first fraction by the second. Did I handle that correctly by multiplying by the reciprocal? Yes, that's the standard approach.Also, I should consider the restrictions on (x). As I noted earlier, (x) cannot be 2, 3, 4, or 5. The simplified expression doesn't include these restrictions explicitly, but they are implied because the original expression was undefined at those points.So, unless there's a step I missed, I think option C is the correct answer. But let me look at the options again:A. (frac{x-1}{x-5})B. (frac{(x-1)(x-5)}{(x-4)(x-2)(x-3)})C. (frac{(x-1)(x-5)}{(x-4)(x-2)})D. (frac{(x-3)(x-5)}{(x-4)(x-2)})Yes, option C is what I arrived at. So, I think that's the correct choice.