🧠:[<think>Okay, let me see. The problem says that the line y = -3x + k does not pass through the third quadrant. I need to find the range of k for which this is true. Hmm, alright. Let me start by recalling some basics about coordinate geometry and quadrants.First, the coordinate plane is divided into four quadrants. The third quadrant is where both the x and y coordinates are negative. So, any point in the third quadrant has x < 0 and y < 0. The line y = -3x + k must not pass through any such point. Therefore, I need to ensure that for all x < 0, the corresponding y value is not also negative. Wait, but maybe there's a better way to approach this.Alternatively, maybe I can analyze the line's path through the quadrants. The line has a slope of -3, which is negative, so the line is decreasing. That means as x increases, y decreases. Now, depending on the y-intercept k, the line will cross the y-axis at (0, k). If k is positive, the line starts high on the y-axis and goes downward as x increases. If k is negative, it starts low on the y-axis and goes downward as x increases.Since the line has a negative slope, it will pass through the second and fourth quadrants for sure, but whether it passes through the first or third depends on the intercept. Wait, no. Let me think. If the slope is negative, the line can pass through quadrants II, I, III, and IV depending on the intercept. Hmm, maybe I need to visualize this.Let me sketch mentally: a line with slope -3. If the y-intercept is positive, then when x is 0, y is positive. As x increases (moves to the right), y decreases, so the line goes from upper left to lower right. So, starting at (0, k) where k is positive, moving to the right, it goes down. Moving to the left (x decreasing), since slope is -3, as x becomes more negative, y increases. So, if k is positive, when x is very negative (left side), y becomes very positive. So the line passes through quadrants II (left, upper), then I (right, upper if it ever gets there), but since slope is negative, when moving to the right, it goes down. Wait, maybe when x is positive enough, y becomes negative. So in that case, the line passes through quadrant IV. So, with positive k, the line passes through quadrants II, I, and IV. So it doesn't pass through III.But wait, if the slope is negative, and the intercept is positive, does it pass through quadrant III? Quadrant III is where x and y are both negative. To pass through quadrant III, the line must have points where x < 0 and y < 0. If the y-intercept is positive, then when x is negative, what happens to y? Let's compute y when x is negative. If x is negative, then -3x becomes positive (because multiplying negative by -3 gives positive). Then y = positive + k. Since k is positive, y is positive plus positive, so y is positive. Therefore, for any x < 0, y is positive. Therefore, the line doesn't pass through quadrant III. So if k is positive, the line doesn't pass through quadrant III.Wait, so if k is positive, then the line is y = -3x + k. For x < 0, y = -3x + k. Since x is negative, -3x is positive, so y is positive plus positive, so y is positive. Therefore, when x is negative, y is positive, so points in quadrant II. When x is positive, y = -3x + k. If x is large enough, y becomes negative, so quadrant IV. So, with positive k, the line passes through II, I (maybe?), and IV. Wait, does it pass through I?When x is between 0 and k/3, y will be positive. For x = 0, y = k. For x = k/3, y = 0. So between x = 0 and x = k/3, the line is in quadrant I. Then for x > k/3, y is negative, quadrant IV. So yes, if k is positive, the line passes through II, I, and IV. But not III.But if k is negative, then the y-intercept is negative. Let's see. Then, when x is 0, y is negative. The slope is -3, so as x increases (moving to the right), y decreases further. So starting at (0, k) where k is negative, moving right, y becomes more negative. Moving left (x decreasing), since slope is -3, for each decrease in x, y increases by 3. So, when x is negative, y = -3x + k. Since x is negative, -3x is positive, so y is positive plus k. If k is negative, then depending on how big |k| is, y might still be positive or negative. For example, if k is -6, then y = -3x -6. When x is -1, y = 3 -6 = -3, which is still negative. So even if x is negative, y might still be negative. Therefore, the line would pass through quadrant III.But if k is negative, then is there a point where x is negative and y is negative? Let's check. Suppose k is negative. Let's take x negative. Then y = -3x + k. Since x is negative, -3x is positive, so y = positive + k. If k is negative, then it's a positive number plus a negative number. Whether y is positive or negative depends on which is larger. If the positive part (-3x) is larger than |k|, then y is positive. If not, y is negative.So, for x < 0, y = -3x + k. Let's solve for when y < 0:-3x + k < 0Since x < 0, let's let x = -a, where a > 0.Then, substituting:-3(-a) + k < 0 => 3a + k < 0 => 3a < -k => a < -k/3But a is positive, so -k/3 must be positive, which implies k < 0. Which is the case here since we are considering k negative.Therefore, when x < 0 (i.e., a > 0), y < 0 when a < -k/3. Therefore, for x in (-(-k/3), 0) = (k/3, 0). Wait, no, hold on. If a < -k/3, then since a = -x, x = -a > -(-k/3) = k/3. But k is negative, so k/3 is negative. So x > k/3, but since k is negative, k/3 is negative. So x is between k/3 and 0. But x is negative here because we are considering x < 0. So x is between k/3 (which is a negative number) and 0. Therefore, in this interval, y < 0. Therefore, the line passes through quadrant III for these x values. Because x is negative (since k/3 is negative and 0), and y is negative.Therefore, if k is negative, the line passes through quadrant III. Hence, in order for the line not to pass through quadrant III, k must be non-negative? Wait, but when k = 0, let's check.If k = 0, the line is y = -3x. Let's see when x is negative, y = -3x. If x is negative, y is positive. So for x < 0, y > 0. So the line is in quadrant II. When x is positive, y is negative, so quadrant IV. So when k = 0, the line passes through quadrants II and IV. So it doesn't pass through III. Therefore, k = 0 is allowed.Wait, but the problem says "the line y = -3x + k does not pass through the third quadrant". So if k = 0, it's okay. If k > 0, also okay. If k < 0, then as shown above, the line does pass through quadrant III. Therefore, the range of k is k ≥ 0? So k is non-negative.But let me verify this. Let me take k = -1. Then the line is y = -3x -1. Let's check when x is negative, say x = -1. Then y = 3 -1 = 2, which is positive. So that point (-1, 2) is in quadrant II. But let's check x = -0.5. Then y = 1.5 -1 = 0.5. Still positive. x = -0.1, y = 0.3 -1 = -0.7. Wait, that's negative. Wait, when x = -0.1, y = -3*(-0.1) + (-1) = 0.3 -1 = -0.7. So x is -0.1 (which is negative), y is -0.7 (negative), so the point (-0.1, -0.7) is in quadrant III. Therefore, even for k = -1, the line passes through quadrant III. Therefore, k must be ≥ 0.But wait, for k = 0, y = -3x. If x is negative, y is positive. If x is positive, y is negative. So it doesn't pass through III. If k is positive, then as we saw, for x < 0, y = -3x + k. Since x is negative, -3x is positive, plus k (positive) gives y positive. So all x < 0 gives y positive, so no points in III. For x positive, y becomes negative once x > k/3, so in IV. So, for k ≥ 0, the line does not pass through III. For k < 0, even though part of the line might be in II and IV, there's a section where x is between k/3 (negative) and 0 where y is negative, hence passing through III.Therefore, the range of k is all real numbers such that k is greater than or equal to 0. So k ≥ 0. Therefore, the answer should be k is in [0, ∞), which in box form is boxed{[0, +infty)}.Wait, but let me check k = 0 again. When k = 0, the line is y = -3x. So when x is 0, y is 0. So the line passes through the origin, which is on the boundary of all quadrants. But technically, quadrants don't include the axes. So the origin isn't part of any quadrant. Therefore, the line y = -3x passes through quadrants II and IV only, right? Because for x > 0, y is negative (IV), and x < 0, y is positive (II). So yes, even k = 0 is okay. So the range is k ≥ 0.Hence, the answer is k ≥ 0. So in box notation, boxed{[0, +infty)}.Final AnswerThe range of ( k ) is boxed{[0, +infty)}.</think>To determine the range of ( k ) such that the line ( y = -3x + k ) does not pass through the third quadrant, we analyze the behavior of the line in different quadrants.1. Understanding the Line's Slope and Intercept: - The line has a slope of -3, which is negative, meaning the line decreases as ( x ) increases. - The y-intercept is ( k ).2. Case Analysis for ( k ): - Positive ( k ): - When ( k ) is positive, the y-intercept is in the positive y-axis. For ( x < 0 ), ( -3x ) is positive, so ( y = -3x + k ) is positive (quadrant II). For ( x > 0 ), ( y ) decreases and eventually becomes negative (quadrant IV). The line does not enter the third quadrant. - Zero ( k ): - When ( k = 0 ), the line is ( y = -3x ). For ( x < 0 ), ( y ) is positive (quadrant II). For ( x > 0 ), ( y ) is negative (quadrant IV). The line passes through the origin, which is not part of any quadrant. - Negative ( k ): - When ( k ) is negative, the y-intercept is in the negative y-axis. For some negative ( x ), ( y = -3x + k ) becomes negative (quadrant III). Thus, the line passes through the third quadrant.3. Conclusion: - The line does not pass through the third quadrant if ( k geq 0 ). For ( k geq 0 ), the line either stays in quadrants II and IV or passes through the origin without entering the third quadrant.Thus, the range of ( k ) is (boxed{[0, +infty)}).,<think>Alright, let's try to figure out this problem. So, the question is: If the line y = -3x + k does not pass through the third quadrant, then what is the range of k? Hmm, okay. Let me start by recalling some basics about lines and quadrants.First, the coordinate plane is divided into four quadrants. The third quadrant is where both x and y are negative. So, if the line doesn't pass through the third quadrant, that means there are no points (x, y) on the line such that x < 0 and y < 0. Got it. So, we need to ensure that for all x < 0, the corresponding y value is not negative. Or maybe there's a different way to approach this.Let me think. The line is given by y = -3x + k. The slope here is -3, which is negative. So, the line is decreasing; as x increases, y decreases. Now, depending on the value of k, the y-intercept, the line will shift up or down. Since the slope is negative, the line will go from the upper right to the lower left if k is positive, or maybe if k is negative, it would start somewhere else. Wait, the y-intercept is (0, k). So, if k is positive, the line crosses the y-axis above the origin; if k is negative, it crosses below.Now, if the line doesn't pass through the third quadrant, we need to make sure that when x is negative (third quadrant), y isn't also negative. But since the line has a negative slope, when x becomes very negative (going left), what happens to y? Let's see. If x is a large negative number, then -3x becomes a large positive number, so y = -3x + k would be a large positive number plus k. So, unless k is a very large negative number, y would still be positive when x is negative. Wait, but if k is negative, then maybe y could be negative even if x is negative?Wait, let's test with some numbers. Suppose k is positive. Let's take k = 5. Then the line is y = -3x + 5. When x is negative, say x = -1, then y = -3*(-1) +5 = 3 +5=8, which is positive. If x is more negative, like x = -10, then y = -3*(-10)+5=30+5=35, still positive. So even with x negative, y is positive. So, if k is positive, then the line doesn't pass through the third quadrant because for all x <0, y is positive. That makes sense because the line crosses the y-axis at (0, k), which is positive, and since it's decreasing, as x increases, y decreases. But when x is negative, it's going to the left from the y-intercept, so the y-values increase. So, if k is positive, the line would pass through quadrants II, I, and IV, but not III. Wait, but quadrant IV is where x >0 and y <0. So, when x is positive, y = -3x + k. If x is big enough, y can be negative, so the line does pass through quadrant IV if k is positive.But the question is about not passing through the third quadrant. So, if k is positive, does the line pass through the third quadrant? From the example above, when k=5 and x is negative, y is positive, so the third quadrant (where both x and y are negative) is not entered. So, k positive is okay.What if k is zero? Then the line is y = -3x. This passes through the origin. So, as x increases, y decreases. When x is positive, y is negative, so quadrant IV. When x is negative, y is positive, quadrant II. So, it passes through quadrants II and IV. But it doesn't pass through III, because when x is negative, y is positive, so not in III. So, k=0 is acceptable.What if k is negative? Let's say k = -2. Then the line is y = -3x -2. Let's check x negative. Let x = -1, then y = -3*(-1) -2 = 3 -2 =1, which is positive. Wait, so x=-1, y=1 is in quadrant II. If x is more negative, like x=-10, then y= -3*(-10) -2=30 -2=28, still positive. So, even with k negative, when x is negative, y is positive? Wait, but hold on. Let me check x=0. Then y= -3*0 + k =k. So, if k is negative, the line crosses the y-axis below the origin. Then, when x increases, y decreases because of the negative slope. So, starting at (0, k) where k is negative, going to the right (increasing x), y becomes more negative. So, that's quadrant IV. Going to the left (decreasing x), the line goes upward because slope is negative. So, when x is negative, y becomes positive as x becomes more negative.But wait, even if k is negative, when x is negative, y = -3x + k. Since x is negative, -3x is positive, so y is positive plus k. If k is negative, but how negative can k be such that y is still not negative when x is negative?Wait a second. Suppose k is negative. Let's take k = -5. Then y = -3x -5. Let's take x = -1. Then y = 3 -5 = -2. So, x=-1, y=-2. That's in the third quadrant! So, when k is negative enough, the line does pass through the third quadrant. Therefore, there must be some threshold value of k where the line just touches the third quadrant or doesn't pass through it. So, our task is to find the range of k such that the line doesn't pass through the third quadrant.So, if k is positive or zero, the line doesn't pass through the third quadrant. But if k is negative, depending on how negative it is, the line may enter the third quadrant. Therefore, there must be a lower bound on k. So, we need to find the values of k for which when x <0, y >=0. So, for all x <0, y = -3x +k >=0. Let me write that inequality.For all x <0, -3x +k >=0.But x is negative, so let's substitute x = -a, where a >0. Then, the inequality becomes:-3*(-a) +k >=0 => 3a +k >=0.Since a >0, 3a is positive. So, 3a +k >=0 must hold for all a >0.But 3a can be made as large as needed by increasing a. Wait, but we need 3a +k >=0 for all a >0. However, if k is negative, then for a sufficiently small, 3a +k might be negative. Wait, but actually, the original condition is that the line does not pass through the third quadrant, which requires that for all x <0, y >=0. But if there exists some x <0 such that y <0, then the line passes through the third quadrant. Therefore, the condition is equivalent to: there does not exist an x <0 such that y <0. So, for all x <0, y >=0. So, for all x <0, -3x +k >=0.But x can be any negative number. Let me analyze the inequality:-3x +k >=0 for all x <0.Let me write this as k >= 3x for all x <0.But x is negative, so 3x is negative. Therefore, the inequality k >= 3x for all x <0.But since x can be any negative number, 3x can be any negative number. So, the right-hand side (3x) can be as small as negative infinity as x approaches negative infinity. Therefore, the inequality k >= 3x for all x <0 would require that k is greater than or equal to all 3x where x is negative. But since 3x can be made arbitrarily negative, the only way k can be greater than or equal to all such 3x is if k is greater than or equal to the supremum of 3x over x <0. However, the supremum of 3x when x <0 is 0, because as x approaches 0 from the negative side, 3x approaches 0. So, the maximum value that 3x can take when x <0 is approaching 0. Therefore, k must be greater than or equal to 0 to satisfy k >= 3x for all x <0. Wait, that seems conflicting with the earlier example where k was negative but for some x <0, y was still positive.Wait, perhaps my approach is incorrect here. Let's try a different angle.To ensure that the line does not pass through the third quadrant, it must not intersect the third quadrant. The third quadrant is where x <0 and y <0. So, if we can find the conditions under which there are no solutions to the system:x <0y = -3x +k <0So, solving for x:From y <0: -3x +k <0 => -3x < -k => 3x >k.But x <0, so 3x >k implies that k <3x. But since x <0, 3x <0. So, this would require that k <0. Therefore, the system x <0 and y <0 is equivalent to k <3x with x <0. Therefore, such x exists if and only if k is less than some negative number. Wait, maybe I need to find the values of k for which there exists an x <0 such that 3x >k. Since x <0, 3x <0. So, 3x >k implies that k <3x <0. Therefore, if k is less than 0, then we can choose x such that k <3x <0. For example, take x = k/3 - ε, where ε is a small positive number. Wait, perhaps not.Alternatively, suppose k is negative. Let's say k = -6. Then 3x >k => 3x > -6 => x > -2. But x has to be <0. So, x is in (-2,0). Therefore, for x between -2 and 0, y would be <0. So, in this case, the line passes through the third quadrant when x is between -2 and 0. Wait, but x is between -2 and 0, which is still negative, and y is negative. So, that's the third quadrant. Therefore, if k is negative, there exists x <0 such that y <0, which is in the third quadrant. Therefore, the line passes through the third quadrant if k is negative. But wait, earlier when I took k=-5 and x=-1, which is less than 0, we had y=-2, which is in the third quadrant. So, this suggests that if k is negative, the line does pass through the third quadrant. But if k is non-negative, then when x is negative, y is positive, so it doesn't pass through the third quadrant. Therefore, the range of k is k >=0.But wait, the initial thought when k=0, the line passes through the origin. The origin is the boundary of all four quadrants. But technically, the origin is not part of any quadrant. So, if the line is y=-3x, which passes through the origin, does it pass through the third quadrant? Since the origin is not in any quadrant, the line y=-3x passes through quadrants II and IV. So, it doesn't pass through III. Therefore, k=0 is acceptable.But when I considered k negative, even slightly negative like k=-1, then the line would cross the y-axis at (0,-1). Then, moving to the left (x decreasing), since the slope is -3, which is negative, going left (x more negative), y increases. So, at x=-1, y = -3*(-1) + (-1) = 3 -1 = 2, which is positive. Wait, that's in quadrant II. Wait, but when x is between 0 and some positive number, y will be negative. For example, x=1, y=-3(1) -1=-4, which is quadrant IV. But to check if it passes through the third quadrant, let's see if there's an x <0 with y <0. Let's try x=-0.5. Then y= -3*(-0.5) + (-1)=1.5 -1=0.5, which is positive. x=-0.1: y=0.3 -1= -0.7. Wait, that's negative. So, when k=-1 and x=-0.1, y=-0.7, which is in the third quadrant. Wait, so that's contradictory to the previous calculation. Wait, let's compute that again. If k=-1, x=-0.1:y= -3*(-0.1) + (-1)=0.3 -1= -0.7. Yes, that's negative. So, x=-0.1 <0, y=-0.7 <0. Therefore, the point (-0.1, -0.7) is in the third quadrant. Therefore, the line y=-3x -1 does pass through the third quadrant. Therefore, even with k=-1, which is not very negative, the line passes through the third quadrant. So, this contradicts my previous thought that if k is negative, but when x is negative, y is positive. It's not the case. When k is negative, even slightly, there are x negative such that y is negative. Therefore, the line passes through the third quadrant.Wait, so how come when k=-5 and x=-1, we had y=3*(-1) + (-5) = -3 -5=-8? Wait, no, wait. Wait, no, the equation is y = -3x +k. If x=-1, then y= -3*(-1) + (-5)=3 -5=-2. So, that's correct. But if k=-1, x=-0.1:y= -3*(-0.1) + (-1)=0.3 -1=-0.7. So, indeed, y is negative. Therefore, even if k is just slightly negative, there exists x negative such that y is negative. Therefore, the line passes through the third quadrant.Therefore, the only way to prevent the line from passing through the third quadrant is to have k >=0. Because if k is non-negative, then when x is negative, y = -3x +k. Since x is negative, -3x is positive, and adding k (which is non-negative) keeps y positive. Therefore, for all x <0, y >=0, so the line doesn't pass through the third quadrant.But let's confirm with k=0. The line is y=-3x. When x=0, y=0. When x is positive, y negative (quadrant IV), when x is negative, y positive (quadrant II). So, doesn't pass through III. If k=0.1, the line crosses the y-axis at (0, 0.1). Moving left (x decreasing), y increases, so stays positive. Moving right (x increasing), y decreases into negative. So, still doesn't pass through III.If k is positive, same thing. So, if k is non-negative, the line doesn't pass through the third quadrant. If k is negative, there are points in the third quadrant. Therefore, the range of k is k >=0.Wait, but the answer is supposed to be a range. The problem says "the range of k is." So, in boxed notation, it would be boxed{[0, +infty)}.But let me check again. Suppose k is zero. The line is y=-3x. At x=0, y=0. For x negative, y positive. For x positive, y negative. So, it passes through II and IV, and touches the origin, which is not a quadrant. So, that's okay. If k is positive, similar situation but shifted up. So, the line starts higher on the y-axis, goes down to the right, passes through I, II, and IV. Wait, when k is positive, does it pass through quadrant I?Yes. For example, take k=5. The y-intercept is (0,5). When x=1, y=-3(1)+5=2, which is still positive. So, in quadrant I. When x is larger than 5/3, y becomes negative. So, it passes through I, II, and IV. But not III. So, even with k positive, the line does not enter III. So, indeed, k >=0 is the correct range.But just to make sure, let me think again. Suppose k=0. Then the line is y=-3x. For x <0, y >0. For x >0, y <0. So, quadrants II and IV. So, no III. If k=1, y=-3x +1. When x=0, y=1. When x=1/3, y=0. So, passes through (1/3,0). For x <0, y= -3x +1. Since x is negative, -3x is positive, so y is positive +1, which is even more positive. So, for x <0, y >1 >0. So, no points in III. For x between 0 and 1/3, y is positive, so quadrant I. For x >1/3, y is negative, quadrant IV. So, passes through I, II, IV. So, no III.If k=-1, as before, when x=-0.1, y=-0.7, which is in III. Hence, for k <0, line passes through III.Therefore, the range of k must be all real numbers greater than or equal to zero. So, the answer is [0, ∞). Therefore, boxed, it's boxed{[0, +infty)}.Final AnswerThe range of ( k ) is boxed{[0, +infty)}.</think>To determine the range of ( k ) such that the line ( y = -3x + k ) does not pass through the third quadrant, we need to ensure that for all ( x < 0 ), the corresponding ( y ) value is non-negative. 1. Understanding the line's behavior: - The line ( y = -3x + k ) has a negative slope (-3) and a y-intercept at ( (0, k) ). - For ( x < 0 ), the term ( -3x ) becomes positive, so ( y = -3x + k ) will be the sum of a positive number and ( k ).2. Analyzing the y-intercept: - If ( k geq 0 ), then for any ( x < 0 ), ( y = -3x + k ) will be positive because both ( -3x ) (positive) and ( k ) (non-negative) contribute to a non-negative result. - If ( k < 0 ), there exists some ( x < 0 ) such that ( y = -3x + k ) will be negative. For example, if ( k = -1 ) and ( x = -0.1 ), then ( y = -3(-0.1) + (-1) = 0.3 - 1 = -0.7 ), which is negative and in the third quadrant.3. Conclusion: - When ( k geq 0 ), the line does not pass through the third quadrant because ( y ) remains non-negative for all ( x < 0 ). - When ( k < 0 ), there are values of ( x < 0 ) such that ( y ) becomes negative, causing the line to pass through the third quadrant.Thus, the range of ( k ) is (boxed{[0, +infty)}).]

🔑:Since the line y = -3x + k does not pass through the third quadrant,- When it passes through the first and fourth quadrants, k = 0,- When it passes through the first, second, and fourth quadrants, k > 0,Therefore, k geq 0.Hence, the answer is: boxed{k geq 0}.

🧠:[<think>Okay, so I need to find the probability that Dr. Zhang is selected when the hospital randomly chooses 2 doctors out of 4, which includes Dr. Zhang. Let me think about how to approach this step by step.First, probability problems often involve figuring out the number of favorable outcomes divided by the total number of possible outcomes. So, maybe I should start by calculating the total number of ways to select 2 doctors from 4. Then, figure out how many of those combinations include Dr. Zhang. Dividing those two should give the probability.Let me recall, the number of ways to choose k items from n without considering the order is called combinations, right? The formula is n choose k, which is n! divided by (k!(n - k)!). So, in this case, total combinations would be 4 choose 2.Calculating that: 4! / (2! * (4 - 2)!) = (4 * 3 * 2 * 1) / (2 * 1 * 2 * 1) = 24 / 4 = 6. So there are 6 possible ways to select 2 doctors out of 4.Now, how many of these combinations include Dr. Zhang? Let's denote the doctors as Dr. Zhang (Z) and three others (A, B, C). If we need to choose 2 doctors and one of them is Z, the other doctor must be one of A, B, or C. So, the possible pairs are (Z, A), (Z, B), (Z, C). That's 3 combinations.Alternatively, using combinations formula again: if one spot is fixed for Dr. Zhang, we need to choose 1 more doctor from the remaining 3. So it's 3 choose 1 = 3. So, 3 favorable outcomes.Therefore, the probability is favorable over total, which is 3/6. Simplifying that gives 1/2. Wait, 3 divided by 6 is 0.5, so 50% chance? That seems high, but maybe it's correct.Wait, let me double-check. The total number of ways to choose 2 out of 4 is 6. The number of pairs that include Dr. Zhang is 3. So yes, 3/6 reduces to 1/2. So the probability is 1/2 or 50%.But let me think differently. What's the probability that Dr. Zhang is selected in the first pick? It's 1/4. Then, given that he wasn't picked first, the probability he's picked second is... Wait, maybe another approach is better.Alternatively, the probability that Dr. Zhang is selected can be thought of as 1 minus the probability that he isn't selected at all. So, total probability is 1 - (number of ways to choose 2 doctors without him divided by total ways).Number of ways without him: choosing 2 from the remaining 3, which is 3 choose 2 = 3. So 1 - 3/6 = 1 - 1/2 = 1/2. So same result.Another way: when selecting 2 people, each person has an equal chance of being selected. There are 4 doctors, each has the same chance. The number of pairs that include a particular doctor is 3 (as we saw), so the probability that Dr. Zhang is in the pair is 3/6 = 1/2.Wait, but is there another way to think about it? For example, the probability that he is selected in the first or the second pick.Probability he is selected first: 1/4. Then, if he is not selected first, the probability he is selected second is 1/3 (since one person has already been selected). So total probability would be (1/4) + (3/4)*(1/3) = 1/4 + 1/4 = 1/2. Yep, same result.So, all approaches lead to 1/2. So the probability is 1/2. But wait, let me check with actual enumeration.Doctors: Z, A, B, C.Possible pairs:1. Z and A2. Z and B3. Z and C4. A and B5. A and C6. B and CSo, out of 6 pairs, 3 include Dr. Z. So yes, 3/6=1/2. That's correct.Therefore, the probability is 1/2.Wait, but intuitively, if you have 4 people and you pick 2, does each person have a 50% chance of being selected? Because each person is in half of the possible pairs. Since each pair combination includes 2 people, and there are 6 pairs, each person is in 3 pairs. 4 people, each in 3 pairs, 4*3=12, but each pair has 2 people, so total is 6*2=12. So yes, each person is in 3 pairs, so 3/6=1/2. So, each individual has a 50% chance of being selected when choosing 2 out of 4. So, that's a general result. If you choose k out of n, the probability that a specific individual is chosen is C(n-1, k-1)/C(n, k). Let me check that formula.Yes, because to count the number of groups containing the specific individual, you fix that person and choose k-1 from the remaining n-1. So, the probability is [C(n-1, k-1)] / C(n, k). Let's apply that here: C(3,1)/C(4,2) = 3/6 = 1/2. Exactly. So, that confirms it.Therefore, regardless of the method, the probability is 1/2. So, the answer is 1/2. But the problem might expect the answer in fraction form, so 1/2 or 50%, but in the box, likely boxed{dfrac{1}{2}}.Wait, but let me check once again if I didn't make a mistake in the setup. The problem says "randomly select 2 doctors from 4 backbone surgeons including Dr. Zhang". So, it's 4 doctors, one of whom is Dr. Zhang. So, all calculations above are correct. So, the answer is 1/2.Alternatively, if there are 4 surgeons total, and selecting 2, then yes. If the problem had more or less, the answer would change, but as per the problem statement, it's 4, so 1/2 is correct.Final AnswerThe probability of selecting Dr. Zhang is boxed{dfrac{1}{2}}.</think>To find the probability of selecting Dr. Zhang when randomly choosing 2 doctors out of 4 (including Dr. Zhang), we start by calculating the total number of ways to select 2 doctors from 4. This is given by the combination formula ( binom{n}{k} ):[binom{4}{2} = frac{4!}{2!(4-2)!} = frac{4 times 3}{2 times 1} = 6]Next, we determine the number of favorable outcomes where Dr. Zhang is selected. If Dr. Zhang is one of the selected doctors, we need to choose 1 more doctor from the remaining 3. This is calculated as:[binom{3}{1} = 3]The probability of selecting Dr. Zhang is the ratio of favorable outcomes to the total number of outcomes:[frac{3}{6} = frac{1}{2}]This result can also be verified by considering the probability that Dr. Zhang is not selected and subtracting it from 1, or by enumerating all possible pairs and checking those that include Dr. Zhang. All methods confirm the probability is ( frac{1}{2} ).Thus, the probability of selecting Dr. Zhang is boxed{dfrac{1}{2}}.,<think>Okay, let's see. The problem is about finding the probability that Dr. Zhang is selected when choosing 2 doctors out of 4, which includes him. Hmm. Probability problems can sometimes be tricky, but let's break it down step by step.First, I need to figure out the total number of possible ways to select 2 doctors from 4. Since the order of selection doesn't matter here—whether you pick Doctor A first or Doctor B first doesn't change the team composition—it's a combination problem. The formula for combinations is C(n, k) = n! / (k!(n - k)!), where n is the total number of items, and k is the number of items to choose.So, plugging in the numbers, the total number of ways is C(4, 2). Let's calculate that. 4! is 24, and 2! is 2. So, 24 / (2 * (4 - 2)!) = 24 / (2 * 2!) = 24 / (2 * 2) = 24 / 4 = 6. So there are 6 possible teams.Now, we need to find how many of these teams include Dr. Zhang. Let's think. If Dr. Zhang is definitely one of the two selected, then we need to choose 1 more doctor from the remaining 3. Because once we've fixed Dr. Zhang as part of the team, we just need to pick one more person to join him.So, the number of favorable outcomes is C(3, 1). Calculating that, it's 3! / (1!(3 - 1)!) = 6 / (1 * 2) = 3. So there are 3 teams that include Dr. Zhang.Therefore, the probability is the number of favorable outcomes divided by the total number of possible outcomes. So that's 3 / 6, which simplifies to 1/2. So the probability is 1/2.Wait, but let me verify this another way to make sure I didn't make a mistake. Alternatively, I can consider each doctor's chance of being selected. Since we're choosing 2 out of 4, each doctor has a 2/4 chance, which is 1/2. So that's another way to see that Dr. Zhang's probability is 1/2. That seems consistent.Alternatively, list all possible combinations to check. Let's name the doctors as A (Dr. Zhang), B, C, D. The possible teams are:1. A & B2. A & C3. A & D4. B & C5. B & D6. C & DOut of these 6 teams, the ones that include Dr. Zhang (A) are the first three: 1, 2, 3. So that's 3 out of 6, which is 1/2. Yep, that matches. So the answer is 1/2.Wait, but another way to think about it: the probability of selecting Dr. Zhang in the first pick is 1/4, but since we are choosing two people, maybe we need to consider both picks. Wait, but combinations don't consider order. So the probability that Dr. Zhang is selected in either the first or the second position.Alternatively, using probability rules. The probability that Dr. Zhang is selected in the first pick is 2/4 (since we are choosing 2 spots out of 4). Wait, no, that might not be the right approach. Let's think step by step.Alternatively, the probability that Dr. Zhang is selected in the first position is 1/4, but since we need to choose two people, perhaps the probability is the sum of the probability that he is picked first plus the probability that he is picked second. Since once someone is picked first, they are not replaced? Wait, but in combinations, it's without replacement. Wait, actually, when you choose two people, you can think of it as picking one person, then another, without replacement. So the probability that Dr. Zhang is picked in either the first or the second selection.So, the probability he is picked first is 1/4, and then after that, the probability he is picked second is 3/4 * 1/3 (since someone else was picked first, then he is picked second). Wait, hold on:Wait, maybe it's easier to calculate the probability that he is not selected, and subtract that from 1.The total number of teams without Dr. Zhang is C(3, 2) = 3. So the probability he is not selected is 3/6 = 1/2. Therefore, the probability he is selected is 1 - 1/2 = 1/2. So that gives the same result. Okay, that seems solid.Alternatively, thinking of the selection as two steps. The first step: probability of selecting Dr. Zhang first is 1/4. Then, since one person is selected, the next selection is 1 out of the remaining 3. But since we need to select two people, the total number of possible selections is 4 * 3 / 2! = 6, which is the same as combinations. Alternatively, if we don't divide by 2!, it's permutations. But in probability, since order doesn't matter, we have to account for that.Alternatively, the probability that Dr. Zhang is selected is equal to the number of selections where he is included, which is 3, divided by total selections 6, so 1/2. So all these different methods give the same answer. Therefore, I feel confident that the probability is 1/2.Final AnswerThe probability of selecting Dr. Zhang is boxed{dfrac{1}{2}}.</think>To find the probability of selecting Dr. Zhang when choosing 2 doctors out of 4, we start by calculating the total number of possible combinations. Using the combination formula ( C(n, k) = frac{n!}{k!(n-k)!} ):[C(4, 2) = frac{4!}{2!(4-2)!} = frac{24}{2 times 2} = 6]Next, we determine the number of combinations that include Dr. Zhang. If Dr. Zhang is one of the selected doctors, we need to choose 1 more doctor from the remaining 3:[C(3, 1) = frac{3!}{1!(3-1)!} = frac{6}{1 times 2} = 3]The probability is the number of favorable outcomes divided by the total number of possible outcomes:[frac{3}{6} = frac{1}{2}]We can verify this by considering the probability of not selecting Dr. Zhang and subtracting from 1. The number of combinations without Dr. Zhang is ( C(3, 2) = 3 ), so the probability of not selecting him is ( frac{3}{6} = frac{1}{2} ). Thus, the probability of selecting him is:[1 - frac{1}{2} = frac{1}{2}]Both methods confirm the probability of selecting Dr. Zhang is (boxed{dfrac{1}{2}}).]

🔑:This problem tests the probability of equally possible events. The relevant knowledge point is that the probability is equal to the ratio of the number of situations sought to the total number of situations.Let's assume that the other three doctors are Dr. Wang, Dr. Li, and Dr. Liu. When randomly selecting two doctors, the equally possible situations that can occur are: (Zhang, Wang); (Zhang, Li); (Zhang, Liu); (Wang, Li); (Wang, Liu); (Li, Liu). There are 6 situations in total.Therefore, P(Selecting Dr. Zhang)= frac{3}{6}= frac{1}{2}.So, the answer is boxed{frac{1}{2}}.

🧠:[<think>Alright, so I need to figure out how many different ways there are to color this map with five administrative areas using four colors. The main rule is that adjacent areas can't share the same color. Hmm, okay, let me break this down step by step.First off, I remember that map coloring problems like this are related to graph theory. Each area is a node, and adjacent areas are connected edges. So, if I can model the map as a graph, I can use graph coloring techniques to solve it. But wait, I don't have the figure mentioned in the problem. Hmm, that's a problem. Maybe I can assume a common configuration? Let me think. Five areas divided into administrative regions... Maybe it's something like a central area surrounded by four others? That seems possible. So, like area 1 in the center, and areas 2, 3, 4, 5 around it, each adjacent to the center and maybe to their neighbors? Or maybe adjacent in a ring?Wait, without the figure, it's a bit hard, but maybe the problem is standard. Let me check my knowledge. If the map is a central region surrounded by four others forming a ring, then the central one (say area 1) is adjacent to all four outer areas (2,3,4,5), and each outer area is adjacent to its two neighbors. So, area 2 is adjacent to 1, 3, and 5; area 3 is adjacent to 1, 2, and 4; etc. Wait, no, if it's a ring, each outer area is adjacent to the center and its two immediate neighbors. So, areas 2 adjacent to 1, 3, and maybe 5? Hmm, maybe not. Wait, in a typical five-region map where one is in the center and four around, each outer region is adjacent to the center and the two adjacent outer regions. So, for example, area 2 is adjacent to 1, 3, and 5; area 3 is adjacent to 1, 2, and 4; area 4 is adjacent to 1, 3, and 5; area 5 is adjacent to 1, 2, and 4. Is that the case? Wait, but if it's a ring, each outer area would be adjacent to two others. Wait, maybe I need to get this straight.Alternatively, maybe the map is like a pentagon with a center. So, the central area is adjacent to all five outer areas, but the problem says five administrative areas. Wait, the problem says "divided into five administrative areas," so maybe it's a central area and four around it? But then that's five total. Wait, no, central plus four would be five. So, central area (1) and four surrounding areas (2,3,4,5). Each surrounding area is adjacent to the center and their immediate neighbors. So, area 2 adjacent to 1, 3, and 5; area 3 adjacent to 1, 2, and 4; area 4 adjacent to 1, 3, and 5; area 5 adjacent to 1, 2, and 4. Hmm, in that case, each outer area is adjacent to three others. But that seems like a possible configuration. But maybe they form a ring without the cross-adjacents? So, area 2 adjacent to 1, 3; area 3 adjacent to 1, 4; area 4 adjacent to 1, 5; area 5 adjacent to 1, 2. But that would make each outer area adjacent only to the center and one other. But that might not be the case. Wait, without the figure, this is challenging.Alternatively, maybe the regions are arranged in a chain? Like 1-2-3-4-5, each adjacent to the next. But that's a linear graph. But with five regions, coloring would be straightforward. But the problem mentions adjacent areas must not use the same color, with four colors. So, if it's a linear graph, the number of colorings would be 4*3*3*3*3 = 4*3^4 = 324. But that seems too high for a problem like this. Wait, but the example 3 is probably a standard one. Wait, perhaps it's a central node connected to all others, and the others form a cycle? So, like a wheel graph with five nodes. The center connected to four others, which form a cycle. So, in graph theory terms, that's called a wheel graph W5. Let me recall: a wheel graph Wn is a graph with n nodes formed by connecting a single central node to all nodes of a cycle of (n-1) nodes. So, W5 would have a center connected to four nodes arranged in a cycle. So, in this case, the four outer nodes form a cycle, each connected to the center. So, each outer node is adjacent to two other outer nodes and the center.In that case, the graph is the wheel graph W5. So, how do you compute the number of colorings for a wheel graph with five nodes and four colors? Let's recall that the chromatic polynomial of a graph gives the number of colorings for a given number of colors. But maybe there's a smarter way than computing the chromatic polynomial.Alternatively, maybe using the principle of multiplication. Let's try to approach it step by step. Let's assume the regions are arranged such that there's a central region (say region 1) connected to four other regions (2,3,4,5), and regions 2,3,4,5 form a cycle (each connected to the next, with 5 connected back to 2). So, that's the wheel graph.If that's the case, then to compute the number of colorings with four colors such that no two adjacent regions share the same color, we can proceed as follows:First, color the central region (1). There are 4 choices.Then, color the surrounding regions (2,3,4,5). Since each of these is adjacent to the center, they can't be the same color as the center. Additionally, each is adjacent to its two neighbors in the cycle.So, after choosing a color for region 1, each of the outer regions must be colored with one of the remaining 3 colors (since they can't be the same as the center). However, the outer regions form a cycle of four nodes. So, the problem reduces to counting the number of colorings of a cycle graph C4 with 3 colors, where adjacent nodes have different colors.Wait, but hold on, if the center is colored with one color, then the outer cycle needs to be colored with the remaining three colors, but adjacent outer regions can't have the same color. However, since the outer regions form a cycle, the number of colorings for a cycle of n nodes with k colors is given by (k-1)^n + (-1)^n (k-1). Wait, is that right? Let me recall the formula for the number of proper colorings of a cycle graph. For a cycle with n nodes, the chromatic polynomial is (k - 1)^n + (-1)^n (k - 1). So, for n=4 and k=3, that would be (3-1)^4 + (-1)^4*(3 - 1) = 2^4 + 1*2 = 16 + 2 = 18. So, 18 colorings for the outer cycle with 3 colors.Therefore, the total number of colorings would be 4 (choices for the center) multiplied by 18 (colorings for the outer cycle), so 4*18=72.But wait, let me verify that. Alternatively, for the outer cycle, since they form a cycle of four nodes, and each adjacent pair must be different colors, and we have 3 colors available.Alternatively, we can model it step by step. Let's suppose we have four outer regions: 2,3,4,5 arranged in a cycle. Let's start with region 2. Since it can't be the same color as the center, it has 3 choices. Then, region 3 can't be the same color as region 2 or the center. Wait, region 3 is adjacent to region 2 and the center. Wait, no, in the wheel graph, region 3 is adjacent to the center and to regions 2 and 4. Wait, hold on. If the outer regions form a cycle, then each outer region is adjacent to two other outer regions and the center. So, for example, region 2 is adjacent to 1 (center), 3, and 5. Wait, no. In a cycle, each outer region is adjacent to two others. If it's a wheel graph, the center is connected to all outer regions, and each outer region is connected to its two immediate neighbors. So, for four outer regions, it's a cycle of four: 2 connected to 3 and 5; 3 connected to 2 and 4; 4 connected to 3 and 5; 5 connected to 4 and 2. Wait, no, with four outer regions, the cycle would be 2-3-4-5-2? But 2-3-4-5 would require a fifth outer region. Wait, no, if there are four outer regions, it's a square: 2-3-4-5-2. Wait, but 2,3,4,5 arranged in a cycle. So, each outer region is connected to two others. So, region 2 is adjacent to 3 and 5; region 3 is adjacent to 2 and 4; region 4 is adjacent to 3 and 5; region 5 is adjacent to 4 and 2. And all of them are connected to the center, region 1.Therefore, when coloring the outer regions, each must be different from the center and different from their two neighbors in the cycle.So, after coloring the center (4 choices), each outer region has 3 available colors (excluding the center's color). However, the outer regions form a cycle where each adjacent pair must also be different.Therefore, the problem reduces to coloring a cycle graph C4 with 3 colors. The number of proper colorings for a cycle of length n with k colors is (k-1)^n + (-1)^n (k-1). For n=4, k=3: (3-1)^4 + (-1)^4*(3-1) = 16 + 2 = 18. So, 18 colorings.Hence, total colorings: 4 * 18 = 72. So, 72 possible colorings.But let me make sure this is correct. Alternatively, perhaps I can compute it manually for the cycle of four nodes with three colors.Let's consider the cycle as nodes A, B, C, D. Each adjacent in the cycle: A-B-C-D-A.Number of colorings where adjacent are different, using three colors.For a cycle with even number of nodes (n=4), the formula is (k-1)^n + (k-1)(-1)^n. Wait, with k=3, n=4: (2)^4 + (2)(1) = 16 + 2 = 18. So that's correct.Alternatively, another way to compute it is:First, color node A: 3 choices.Then, color node B: must be different from A: 2 choices.Color node C: must be different from B. If B is different from A, then C can be either A's color or another color. Wait, but since it's a cycle, we have to consider the constraint that D must be different from C and A.Wait, maybe inclusion-exclusion. Let's think of linear arrangements and then adjust for the cycle.If it were a chain (not a cycle), the number of colorings would be 3*2*2*2 = 24. Because first node 3 choices, each subsequent node 2 choices (different from previous). But since it's a cycle, the first and last nodes are adjacent, so we need to subtract the cases where the first and last nodes are the same color.But how to compute that?Number of colorings for a cycle = number of colorings for a chain where the first and last nodes are different.So, total colorings for a chain: 3*2*2*2 = 24.Number of colorings where first and last are the same: Let's compute that.If A and D are the same color. Let's fix A's color (3 choices). Then D must be the same as A. Then B must be different from A: 2 choices. C must be different from B: 2 choices. D is fixed, but must be different from C. Since D is same as A, so C must be different from A. But C is already different from B, which is different from A. Wait, if B is different from A, and C is different from B, then C could be equal to A or another color. Wait, but since D is fixed as A's color, and D must be different from C. Therefore, C must not be A. Therefore, once A is chosen, B has 2 choices, C must be different from B and not equal to A, so if B is not A, then C can be either of the remaining two colors, but one of them is A. Wait, no, since we have three colors. If B is different from A (2 choices), then C must be different from B. If B is color 2, then C can be color 1 or 3. But color 1 is A's color, so if C is color 1, then D (which is color 1) would be adjacent to C, which is also color 1, which is invalid. Therefore, C cannot be color 1. Therefore, C must be color 3 (assuming B is color 2). So, in general, once A is fixed, B has 2 choices, C has 1 choice (the color different from B and A). Then D is fixed as A. So, the number of colorings where A and D are the same is 3 (choices for A) * 2 (choices for B) * 1 (choice for C) = 6.Therefore, the number of colorings where A and D are the same is 6.Therefore, the number of colorings for the cycle is total colorings for chain minus colorings where A and D are the same: 24 - 6 = 18. Which matches the formula. So, yes, 18.Therefore, going back, the total number of colorings is 4 * 18 = 72.But let me check if this is the standard answer. Wait, the problem says "adjacent areas must not use the same color". So, if the regions are arranged as a wheel graph, then the answer is 72. But is this the case? The problem states "a region is divided into five administrative areas as shown in the figure". Since I don't have the figure, I have to assume the most probable configuration. Wheel graph is a common one for such problems. Alternatively, another configuration.Wait, another possible configuration is that the five areas form a pentagon, each adjacent to two others. So, a cycle of five nodes. But then, the number of colorings with four colors would be different. For a cycle of five nodes, the number of colorings with k colors is (k-1)^n + (-1)^n (k-1). For n=5, k=4: (3)^5 + (-1)^5*(3) = 243 - 3 = 240. But with four colors, maybe? Wait, no, the formula is for k colors. Wait, the chromatic polynomial for a cycle graph is (k - 1)^n + (-1)^n (k - 1). Wait, no, that might not be correct.Wait, the chromatic polynomial for a cycle graph Cn is (k - 1)^n + (-1)^n (k - 1). So, for n=5, k=4, it would be (3)^5 + (-1)^5*3 = 243 - 3 = 240. So, 240 colorings. But if the graph is a pentagon, and the problem allows four colors, then the answer would be 240. But the problem mentions four colors, and the example is asking for the number of different coloring methods. Wait, but the example is labeled as Example 3, so maybe it's a standard problem.Wait, but in the initial problem statement, it's divided into five areas, with four colors, adjacent areas different colors. The key is whether the graph is a wheel or a cycle. If it's a wheel, the answer is 72; if it's a cycle, the answer is 240. But given that it's five areas, a wheel makes sense (central area and four surrounding), but without the figure, it's ambiguous.Alternatively, another common five-region map is the one similar to a plus sign, with four regions around a central one. In that case, the central region is adjacent to all four outer regions, and the outer regions are only adjacent to the center, not to each other. If that's the case, the graph is a star graph, K1,4. Then, the number of colorings would be calculated as follows: color the center (4 choices), then each outer region can be colored with 3 choices (different from the center). Since the outer regions are not adjacent to each other, they can have the same color. Therefore, 4 * 3^4 = 4 * 81 = 324. But that seems high. However, the problem says "adjacent areas must not use the same color". If the outer regions are not adjacent to each other, then their colors only need to differ from the center. So, if the graph is K1,4, then 324 colorings. But that contradicts the earlier answer if the graph is a wheel.Therefore, the problem is ambiguous without the figure. However, given that the problem is in Chinese (original mentions "administrative areas" which is common in Chinese problems), and in some Chinese textbooks, the five-region map coloring problem with four colors often refers to a central region surrounded by four regions, each adjacent to the center and to two neighboring outer regions, forming a wheel graph. Therefore, leading to 72 colorings. But I need to confirm.Alternatively, maybe the problem is similar to the one in the textbook where the answer is 72. But let me check another way.Suppose the regions are arranged such that region 1 is adjacent to 2, 3, 4, 5; and regions 2,3,4,5 are each adjacent to the next one, forming a cycle. So, 2 adjacent to 3, 3 adjacent to 4, 4 adjacent to 5, 5 adjacent to 2. So, that's the wheel graph. Then, as above, the calculation would be 4 * 18 = 72.Alternatively, if the regions are arranged such that regions 2,3,4,5 form a square, each adjacent to two others, and all adjacent to the central region 1. Then, indeed, the calculation is 4 * 18 = 72.Alternatively, if the problem had the outer regions not forming a cycle, then the number would be different. For example, if outer regions are only adjacent to the center, then it's 4 * 3^4 = 324. If outer regions form a different structure, the answer changes.But since the problem mentions "adjacent areas must not use the same color" without further details, and given that it's divided into five administrative areas, which often in such problems are designed to have each outer region adjacent to two others and the center, forming a wheel, leading to the 72 answer.Therefore, my conclusion is that the number of different coloring methods is 72. So, the answer is boxed{72}.Wait, but hold on. Let me check once more. Suppose the graph is the wheel graph W5. The chromatic polynomial of a wheel graph can be computed. The chromatic polynomial for Wn is k*(k-2)^{n-1} + (-1)^n*k*(k-2). Wait, maybe not. Alternatively, the chromatic polynomial for a wheel graph can be calculated by considering it as the join of a central node and a cycle graph. The chromatic polynomial of the join of two graphs G and H is P(G, k) * P(H, k) / k, but I might be misremembering.Alternatively, since the wheel graph W5 is a central node connected to a cycle C4. The chromatic polynomial can be calculated as follows. For the central node, we have k choices. For the cycle C4, which is adjacent to the central node, each node in the cycle can't take the central color. Therefore, the chromatic polynomial would be k * chromatic polynomial of C4 with (k-1) colors. So, chromatic polynomial of W5 is k * [( (k - 2)^4 + (k - 2) )]. Wait, because the chromatic polynomial of C4 is (x - 1)^4 + (-1)^4*(x - 1) = x^4 - 4x^3 + 6x^2 - 3x. Wait, no. Wait, actually, the chromatic polynomial for cycle Cn is (x - 1)^n + (-1)^n (x - 1). For n=4, that's (x - 1)^4 + (x - 1) = (x^4 - 4x^3 + 6x^2 - 4x + 1) + (x - 1) = x^4 -4x^3 +6x^2 -3x. So, substituting x with (k - 1), since the cycle can't use the central color. Wait, no. Let's think again. If the central node is colored with one color, then each node in the cycle has (k - 1) choices, but with the restriction that adjacent nodes are different. So, the chromatic polynomial for the cycle with (k - 1) colors. Therefore, substituting x = k - 1 into the chromatic polynomial of C4.The chromatic polynomial of C4 is x(x - 1)^3 + (-1)^4 x = x^4 - 3x^3 + 3x^2 - x + x = x^4 - 3x^3 + 3x^2. Wait, no. Wait, actually, another formula: for cycle Cn, the number of colorings is (k - 1)^n + (-1)^n (k - 1). So, as before, substituting k with (x), number of colorings is (x - 1)^n + (-1)^n (x - 1). So, for n=4, substituting x = k - 1, then the number of colorings for the cycle is ((k - 1) - 1)^4 + (-1)^4*((k - 1) - 1) = (k - 2)^4 + 1*(k - 2) = (k - 2)^4 + (k - 2).Therefore, the total chromatic polynomial for the wheel graph would be k * [ (k - 2)^4 + (k - 2) ].Then, substituting k=4, we get 4 * [ (4 - 2)^4 + (4 - 2) ] = 4 * [ 16 + 2 ] = 4 * 18 = 72. Exactly as before. Therefore, this confirms that for the wheel graph W5 with 4 colors, the number of colorings is 72.Therefore, the answer is 72.However, let me just make sure that if the graph was different, like for example, five regions in a straight line, or some other configuration, the answer would be different. But given that the problem mentions "adjacent areas", and the common pedagogical problem is the wheel graph, I think 72 is the correct answer here.Final Answerboxed{72}</think>To solve the problem of coloring a map divided into five administrative areas with four colors such that adjacent areas do not share the same color, we consider the regions as a wheel graph (W5). This graph consists of a central region (1) connected to four outer regions (2, 3, 4, 5), which form a cycle.1. Coloring the Central Region: There are 4 choices for coloring the central region (1).2. Coloring the Outer Regions: The four outer regions form a cycle (C4) and must be colored with the remaining 3 colors (since they cannot be the same as the central region). The number of ways to color a cycle graph (C4) with 3 colors is calculated using the formula for the chromatic polynomial of a cycle graph: [ (k-1)^n + (-1)^n (k-1) ] For ( n = 4 ) (cycle length) and ( k = 3 ) (remaining colors): [ (3-1)^4 + (-1)^4 (3-1) = 2^4 + 1 cdot 2 = 16 + 2 = 18 ]3. Total Coloring Methods: Multiply the number of choices for the central region by the number of valid colorings for the outer cycle: [ 4 times 18 = 72 ]Thus, the number of different coloring methods is (boxed{72}).,<think>Okay, so I have this map coloring problem. The region is divided into five administrative areas, and we need to color them using four colors. The rule is that adjacent areas can't share the same color. Hmm, I remember something about graph coloring here. Maybe each area is a node, and edges connect adjacent areas? Then the number of colorings would be the chromatic polynomial of that graph evaluated at 4. But wait, I need to make sure how exactly the areas are arranged. The problem mentions a figure, but since there's no figure here, I need to assume a common configuration for five areas. Maybe it's like a central area surrounded by four others? That would make it similar to a wheel graph with five nodes. Let me visualize that: area 1 in the center connected to areas 2, 3, 4, 5 around it, and each of those outer areas are adjacent to their neighbors? So areas 2 and 3 are adjacent, 3 and 4, 4 and 5, and 5 and 2? If that's the case, then the graph is a wheel graph with five vertices.Alternatively, maybe the areas are arranged in a different way. For example, if it's a pentagon where each area is adjacent to two others, plus some other connections. But without the figure, it's hard to tell. But since the example is numbered as Example 3, maybe it's a standard problem. Wait, maybe it's similar to the standard four-color theorem examples. Let me think. If there are five areas arranged such that each adjacent to the next, forming a cycle, but then with one more area connected to all others? That would be a wheel graph. If that's the case, then how do we calculate the number of colorings?Alternatively, maybe the map is a simple five-node graph where each area is connected to some others. Wait, but the problem says "adjacent areas must not use the same color," so we need to model this as a graph coloring problem. The key is knowing the adjacency, but without the figure, I need to make an assumption. Maybe the problem refers to a standard configuration, like the one in the textbook where a central region is surrounded by four others, each adjacent to the center and to their two neighbors. That is, a central area (1) connected to 2, 3, 4, 5, and each of 2,3,4,5 forms a cycle. So areas 2-3-4-5-2. If that's the case, then the graph is a wheel with five nodes. Let me confirm.In graph theory, the chromatic polynomial for a wheel graph can be calculated. The wheel graph Wn is a graph formed by connecting a single universal vertex to all vertices of a cycle of (n-1) vertices. So for n=5, W5 has a center connected to a cycle of four. The chromatic polynomial for a wheel graph Wn is known, but I need to recall the formula. Alternatively, I can compute it step by step.Alternatively, maybe I can use deletion-contraction. But that might be complicated. Alternatively, since we have four colors, and the chromatic polynomial evaluated at k=4 will give the number of colorings. Let's see.First, let's consider the structure: a center area (say area 1) connected to areas 2,3,4,5. Areas 2,3,4,5 form a cycle (each adjacent to the next). So area 2 adjacent to 3, 3 to 4, 4 to 5, 5 to 2, and all adjacent to 1.To compute the number of colorings with 4 colors where adjacent areas have different colors, we can use the principle of multiplication, considering the order of coloring.Alternatively, use the chromatic polynomial formula. The chromatic polynomial for a wheel graph W5. Wait, for a wheel graph with n vertices, the chromatic polynomial is (k - 2)^{n - 1} + (-1)^{n - 1} (k - 2). But I need to verify this. Let me check for n=4. A wheel graph with 4 nodes is a triangle with a central node connected to all three. The chromatic polynomial for W4 is k*(k-1)*(k-2)^2. Wait, maybe the formula is different. Alternatively, since the wheel graph is a cone over a cycle. So for Wn, which is a cycle Cn-1 plus a central vertex connected to all vertices on the cycle. The chromatic polynomial can be calculated by considering the cycle and the central vertex.Alternatively, here's an approach. Let's first color the center area. There are 4 color choices for area 1. Then, we need to color the surrounding areas 2,3,4,5, each of which must be different from area 1 and also different from their adjacent neighbors on the cycle.So after coloring area 1 with one color, the remaining four areas form a cycle where each must be colored with the remaining 3 colors (since they can't be the same as area 1). Wait, no: the problem states there are 4 colors available. So if area 1 is colored with color A, then each of areas 2,3,4,5 can be colored with any color except A. So they have 3 choices each. However, adjacent areas on the cycle (2-3, 3-4, 4-5, 5-2) must also be different from each other. So we need the number of proper colorings of a cycle graph C4 with 3 colors.Wait, no. Wait, each of the outer areas (2,3,4,5) can use any color except the one used by area 1. So if area 1 is colored with one of the 4 colors, then each of the outer areas has 3 choices. However, they also must differ from their adjacent outer areas. So the problem reduces to coloring a cycle of 4 nodes with 3 colors. Wait, but the number of colors available is 3 for each node (excluding the center's color). So the number of colorings for the cycle is the number of proper colorings of C4 with 3 colors. The formula for the number of colorings of a cycle Cn with k colors is (k-1)^n + (-1)^n (k-1). For n=4, it's (k-1)^4 + (k-1). So substituting k=3, we get (2)^4 + 2 = 16 + 2 = 18. Therefore, for each color choice of the center, the number of colorings for the outer cycle is 18. Since the center has 4 color choices, total colorings would be 4 * 18 = 72. But wait, hold on. Wait, no. Wait, the formula for the number of proper colorings of a cycle graph with n nodes and k colors is (k-1)^n + (-1)^n (k-1). So for n=4, it's (k-1)^4 + (k-1). If k=3, then (3-1)^4 + (3-1) = 16 + 2 = 18. So yes, that's correct. So if the outer cycle has 18 colorings for each center color, then total colorings would be 4 * 18 = 72. However, wait, but is that correct?Wait, let's check the formula again. Alternatively, the chromatic polynomial of a cycle Cn is (k - 1)^n + (-1)^n (k - 1). So when evaluated at k=3, it gives the number of colorings. So yes, that gives 18. Therefore, the total colorings would be 4 * 18 = 72. However, is this correct? Let me test with a simpler case.Suppose the center is colored with color A. Then the four outer regions must be colored with B, C, D. The cycle of four nodes with three colors. Let's compute the number of colorings manually for a cycle of four nodes with three colors. For a cycle with four nodes, the number of proper colorings is 3 * 2 * 2 * 2 - ... Wait, maybe not. Alternatively, using permutations. The number of colorings where adjacent colors are different. For a cycle of 4 nodes, the number is (3-1)^4 + (-1)^4*(3-1) = 16 + 2 = 18. Alternatively, using recurrence relations. For a cycle of n nodes, it's (k-1)^n + (-1)^n (k-1). So that's 18. So that formula gives 18, so 4 * 18 = 72. So the answer would be 72. But wait, let me check if there's a mistake here.Wait, another way to compute the number of colorings for the outer cycle. Let's fix the center color as color 1. Then, each outer node can be colored with 2,3,4. The outer nodes form a cycle. So the number of colorings is the number of proper colorings of a 4-node cycle with 3 colors. Let's compute that.For a cycle of 4 nodes with 3 colors, the number of colorings is calculated as follows. The first node has 3 choices. The second node, adjacent to the first, has 2 choices. The third node, adjacent to the second, has 2 choices (it can't be the same as the second). The fourth node is adjacent to both the third and the first. So if the third node is different from the first, then the fourth node has 1 choice. But if the third node is the same as the first, then the fourth node has 2 choices. Wait, this seems complicated. Let's use inclusion-exclusion.Alternatively, the number of colorings is equal to (k-1)^n + (-1)^n (k-1) when the cycle is even or odd. Wait, for n=4 and k=3, it's (3-1)^4 + (-1)^4*(3-1) = 16 + 2 = 18. So that's correct. Therefore, the number is indeed 18. Therefore, the total colorings would be 4 * 18 = 72. So is that the answer? Wait, but let me think again. Suppose the center is colored with one color, then the outer cycle is colored with 3 colors, giving 18 possibilities. So 4 choices for the center, 18 for the cycle, 4*18=72. So that would be the answer. But wait, wait, but what if the outer cycle can sometimes have the same color as the center? Wait, no, because the outer areas are adjacent to the center, so they must be different from the center. Therefore, once the center is colored, the outer areas must be colored with the remaining 3 colors, and adjacent outer areas must be different from each other. Therefore, the outer cycle is effectively being colored with 3 colors, with adjacent ones different. So the formula applies here, giving 18 colorings for the outer cycle per center color. Therefore, 4*18=72. So the answer is 72. But wait, let me confirm with another method.Alternatively, use the chromatic polynomial for the wheel graph. The chromatic polynomial of a wheel graph Wn can be computed as follows: For Wn, which has n vertices, the chromatic polynomial is (k - 2)(k - 1)^{n - 2} + (-1)^{n - 1}(k - 2)(k - 1). Wait, not sure. Alternatively, let's construct it. The wheel graph W5 is formed by connecting a central node to all nodes of a 4-cycle. So the chromatic polynomial is equal to the chromatic polynomial of the 4-cycle multiplied by (k - 1), since the central node is connected to all four nodes. Wait, no. Actually, the chromatic polynomial of a graph G with a node v connected to all other nodes is more complex. Let me recall that if a graph is formed by connecting a new vertex to all vertices of an existing graph H, then the chromatic polynomial is (k - s) * chromatic polynomial of H evaluated at (k - 1), where s is the number of colors? Hmm, maybe not exactly. Alternatively, use deletion-contraction.Wait, perhaps an easier way. Let me consider the total chromatic polynomial for the wheel graph W5. The chromatic polynomial for a wheel graph with 5 nodes can be calculated as follows. Let me denote the central node as C and the cycle nodes as A, B, D, E (connected as A-B-D-E-A). The edges are C connected to A, B, D, E; and the cycle edges A-B, B-D, D-E, E-A.Wait, actually, to compute the chromatic polynomial, maybe use the formula for graphs constructed by joining two graphs. But perhaps it's too involved. Alternatively, let me use the principle of inclusion-exclusion.Alternatively, let's think step by step. First, color the central node C. There are k choices. Then, color the cycle nodes A, B, D, E such that each is different from C and from their adjacent nodes. So once C is colored, each of A, B, D, E has (k - 1) choices, but with the constraint that adjacent nodes on the cycle are different. Therefore, the chromatic polynomial would be k * chromatic polynomial of the cycle C4 evaluated at (k - 1). The chromatic polynomial of a cycle C4 is (k - 1)^4 + (-1)^4*(k - 1) = (k - 1)^4 + (k - 1). Therefore, substituting (k - 1) into this polynomial, we get:Chromatic polynomial of W5 is k * [ ( (k - 1) - 1 )^4 + ( (k - 1) - 1 ) ].Wait, no. Wait, the chromatic polynomial of the cycle C4 is (x - 1)^4 + (-1)^4*(x - 1) where x is the number of colors. Therefore, if we need to substitute x = (k - 1) because each node on the cycle has (k - 1) available colors (since they can't be the same as the center), then the chromatic polynomial for the cycle part is ( (k - 1) - 1 )^4 + (k - 1 - 1 ) = (k - 2)^4 + (k - 2). Therefore, the total chromatic polynomial for the wheel graph is k * [ (k - 2)^4 + (k - 2) ].Therefore, evaluating at k=4, we get 4 * [ (4 - 2)^4 + (4 - 2) ] = 4 * [ 16 + 2 ] = 4 * 18 = 72. Therefore, the number of colorings is 72. So that matches the previous result. Therefore, the answer should be 72. But let me check once more to be sure.Alternatively, suppose we try to compute it step by step. Let's fix the color of the center. There are 4 choices. Then, we need to color the four outer regions with 3 colors, such that adjacent ones are different. For a cycle of four nodes with three colors, the number of colorings is 3 * 2 * 2 * 2 - ... Wait, let's think.First, choose a color for the first outer node (say node A). It has 3 choices (since it can't be the center's color). The next node (B) has 2 choices (can't be the same as A). The next node (D) has 2 choices (can't be the same as B). The fourth node (E) is adjacent to both D and A. So if D and A are different colors, E has 1 choice (can't be D or A). If D and A are the same color, E has 2 choices (can't be D=A or the center's color). Wait, but in our case, A and D are two nodes apart on the cycle. Wait, in a cycle A-B-D-E-A, so A is adjacent to B and E, B is adjacent to A and D, D is adjacent to B and E, E is adjacent to D and A. Therefore, when coloring E, it must be different from D and A. So the color of E depends on whether A and D are the same or different.Therefore, let's compute the number of colorings where A and D are the same color vs different.Case 1: A and D are the same color. Then:- Choose color for A: 3 choices.- Color B: must be different from A, so 2 choices.- Color D: must be same as A, so 1 choice.- Color E: must be different from D (which is A) and E is adjacent to D and A (same color). So E must be different from A, so 2 choices (since the center's color is already excluded, but wait, E can be colored with any color except the center's color and the adjacent nodes. Wait, the center's color is fixed. So E is adjacent to D (which is A) and to the center. So E cannot be the same as D (A) or the center. However, since the center is already a different color, and E is being colored with the remaining 3 colors (excluding the center). So E must be different from D (A) and from the center. But since the center is already a different color, E just needs to be different from D (A). But E is adjacent to D, so E must be different from D. Since D is colored with A (which is not the center's color), then E can be colored with any color except A and the center's color. Wait, but the center's color is already excluded from all outer nodes. Wait, no: the outer nodes cannot be the same as the center's color, right? Wait, the problem states that adjacent areas must not use the same color. So each outer area is adjacent to the center, so they must be different from the center's color. Therefore, when coloring the outer nodes, they can only use the remaining 3 colors (excluding the center's color). So in this case, E must be different from D (which is color A, one of the 3 available) and different from the center's color (which is already excluded). Wait, actually, since all outer nodes are already restricted to the 3 colors not used by the center, the adjacency between outer nodes only needs to consider those 3 colors. Therefore, when coloring E, it's adjacent to D and A. So E must be different from D and A. But since A and D are the same color in this case, E must be different from that color. Therefore, E has 2 choices (since there are 3 colors total, minus 1 for the color of A/D). So in this case:Total colorings for Case 1: 3 (A) * 2 (B) * 1 (D) * 2 (E) = 12.Case 2: A and D are different colors.- Choose color for A: 3 choices.- Color B: 2 choices (different from A).- Color D: must be different from B and also different from A. Since B is already different from A, D has (3 - 2) = 1 choice (if B is different from A, and D must be different from B and A). Wait, hold on. If there are 3 colors available, and D must be different from B and A, which are two different colors (since A and B are adjacent, so different), then D has 1 color left. So D is determined.- Then, color E: adjacent to D and A. Since D and A are different, E must be different from both. Therefore, E has (3 - 2) = 1 choice.So in this case:Total colorings for Case 2: 3 (A) * 2 (B) * 1 (D) * 1 (E) = 6.Wait, but this seems like only 6 colorings for Case 2. But 12 + 6 = 18, which matches the previous result. Therefore, total colorings for the outer cycle is 18 per center color. Therefore, 4 * 18 = 72. Therefore, the answer is 72.But wait, in the Case 2 calculation, after choosing A (3 choices), B (2 choices), then D must be different from B and A. Since there are 3 colors available, and D can't be B or A, so only 1 choice. Then E is adjacent to D and A, which are different colors, so E must be the remaining color. So E has 1 choice. Therefore, 3 * 2 * 1 * 1 = 6. So total for Case 1 and Case 2: 12 + 6 = 18. Therefore, per center color, 18 colorings. 4 center colors gives 4 * 18 = 72. Therefore, the answer is 72.But wait, let me check with another approach. Suppose instead of separating into cases, use permutations. The number of proper colorings of a cycle with 4 nodes and 3 colors is 3 * 2 * 2 * 2 - ... No, perhaps not. Alternatively, the formula (k-1)^n + (-1)^n (k-1) when k=3, n=4 gives 2^4 + 2 = 18. So yes, that's the same.Therefore, the conclusion is 72. Therefore, the number of different coloring methods is 72. So the answer is boxed{72}.Final AnswerThe number of different coloring methods is boxed{72}.</think>To solve the problem of coloring a region divided into five administrative areas with four colors such that adjacent areas do not share the same color, we model the problem using graph theory. The regions form a wheel graph ( W_5 ), which consists of a central area (node) connected to four surrounding areas (nodes) arranged in a cycle.1. Coloring the Central Area: - The central area can be colored in 4 ways (using any of the 4 colors).2. Coloring the Surrounding Cycle: - After coloring the central area, the four surrounding areas must be colored using the remaining 3 colors. - The surrounding areas form a cycle graph ( C_4 ). The number of ways to color a cycle graph ( C_n ) with ( k ) colors is given by the formula: [ (k-1)^n + (-1)^n (k-1) ] - For ( n = 4 ) (a cycle of four nodes) and ( k = 3 ) (remaining colors), the number of colorings is: [ (3-1)^4 + (-1)^4 (3-1) = 2^4 + 2 = 16 + 2 = 18 ]3. Total Coloring Methods: - Multiply the number of ways to color the central area by the number of ways to color the surrounding cycle: [ 4 times 18 = 72 ]Thus, the number of different coloring methods is (boxed{72}).]

🔑:Analysis one: Starting from the 1st position, 1, 2, and 3 have 4, 3, and 2 ways of coloring, respectively. (1) If 4 and 2 are the same color, then 4 has only 1 way of coloring, and 5 has 2 ways, i.e., 4 times 3 times 2 times 1 times 2 = 48 ways; (2) If 4 and 2 are different colors, then 4 has 1 way, and 5 also has 1 way, i.e., 4 times 3 times 2 times 1 times 1 = 24 ways. In total, there are 48 + 24 = 72 ways of coloring.Analysis two: 5 regions with 4 colors, at least two regions must be the same color.(1) If only 3 and 5 are the same color, there are mathrm{C}_{4}^{1} ways of coloring, and 1, 2, 4 being different colors have mathrm{A}_{3}^{3} ways of coloring, totaling mathrm{C}_{4}^{1} cdot mathrm{A}_{3}^{3} = 24 ways of coloring;(2) If only 2 and 4 are the same color, there are mathrm{C}_{4}^{1} ways of coloring, and 1, 3, 5 being different colors have mathrm{A}_{3}^{3} ways of coloring, totaling mathrm{C}_{4}^{1} cdot mathrm{A}_{3}^{3} = 24 ways of coloring;(3) If 2 and 4 are the same color and 3 and 5 are the same color, there are mathrm{A}_{4}^{2} ways of coloring, and 1 has mathrm{C}_{2}^{1} ways of coloring, totaling mathrm{A}_{4}^{2} cdot mathrm{C}_{2}^{1} = 24 ways of coloring.In summary, there are 24 + 24 + 24 = 72 ways of coloring.

🧠:[<think>Okay, let's try to figure out who committed the crime among Brown, Jones, and Smith. Each of them made two statements, and we know that one person told the truth twice, another lied twice, and the third was truthful once and lied once. Hmm, this seems like a logic puzzle where we need to check each possibility and see which one fits the given conditions.First, let me list out all the statements:- Brown says: 1. "I didn't do it." 2. "Jones didn't do it."- Smith says: 1. "I didn't do it." 2. "Brown did it."- Jones says: 1. "Brown didn't do it." 2. "Smith did it."We need to determine who is the criminal such that among the three, one is all truthful (both statements true), one is all lying (both statements false), and one is mixed (one true, one false).Let me approach this by assuming each person as the criminal and checking if the statements fit the required truth-telling conditions.Case 1: Suppose Brown is the criminal.If Brown did it, let's evaluate each person's statements.- Brown's statements: 1. "I didn't do it." → Lie (since he did it) 2. "Jones didn't do it." → True (since Brown did it) So Brown has one lie and one truth. That would make him the mixed one.- Smith's statements: 1. "I didn't do it." → True (Smith is innocent) 2. "Brown did it." → True (since we assumed Brown is guilty) Both statements are true. So Smith would be the all-truthful one.- Jones's statements: 1. "Brown didn't do it." → Lie (Brown did it) 2. "Smith did it." → Lie (Smith didn't do it) Both statements are lies. So Jones is the all-lying one.In this case, the roles fit: Brown (1 lie, 1 truth), Smith (both truths), Jones (both lies). But wait, the problem states that one told the truth twice, another lied twice, and the third told the truth once and lied once. Here, that's exactly the case. So this seems possible. But let's check other cases to be sure.Case 2: Suppose Smith is the criminal.If Smith did it, let's evaluate each person's statements.- Smith's statements: 1. "I didn't do it." → Lie (he did it) 2. "Brown did it." → Lie (Brown didn't do it) So Smith would be the all-lying one.- Brown's statements: 1. "I didn't do it." → True (innocent) 2. "Jones didn't do it." → True (innocent, since Smith did it) Both statements true. Brown is all-truthful.- Jones's statements: 1. "Brown didn't do it." → True (correct, Brown is innocent) 2. "Smith did it." → True (correct, Smith is guilty) Both statements true. But this contradicts the conditions because we can't have two people being all-truthful. So Smith being the criminal leads to two all-truthfuls (Brown and Jones), which isn't allowed. Therefore, Smith cannot be the criminal.Case 3: Suppose Jones is the criminal.If Jones did it, let's evaluate the statements.- Jones's statements: 1. "Brown didn't do it." → True (Brown is innocent) 2. "Smith did it." → Lie (Smith is innocent) So Jones has one truth and one lie. He would be the mixed one.- Brown's statements: 1. "I didn't do it." → True (he's innocent) 2. "Jones didn't do it." → Lie (Jones is guilty) Brown has one truth and one lie. Also mixed. But we can only have one mixed person. Conflict here.- Smith's statements: 1. "I didn't do it." → True (innocent) 2. "Brown did it." → Lie (Brown is innocent) Smith also has one truth and one lie. So now both Brown and Smith are mixed, and Jones is mixed. That's three mixed, which violates the condition. Therefore, Jones cannot be the criminal.Wait, maybe I made a mistake here. Let me check again.If Jones is the criminal:- Jones's statements are 1 true, 1 lie (as above).- Brown's first statement: "I didn't do it." True. Second: "Jones didn't do it." False. So Brown is mixed.- Smith's first statement: "I didn't do it." True. Second: "Brown did it." False. So Smith is mixed.So all three would be mixed, which is impossible. Hence, Jones can't be the criminal.So, from the three cases, only when Brown is the criminal do we satisfy the conditions: Smith is all-truthful, Jones is all-lying, and Brown is mixed.Wait, but earlier, when assuming Brown is guilty, Brown's statements: first is lie, second is truth. So Brown is mixed. Smith's both statements are true (since Smith said "I didn't do it" which is true, and "Brown did it" which is true). Jones's both statements are lies ("Brown didn't do it" is lie, "Smith did it" is lie). So that works: one truthful (Smith), one liar (Jones), one mixed (Brown).Therefore, Brown is the criminal. But let me just check again because sometimes these puzzles can be tricky.Wait a second, but Smith's second statement is "Brown did it," which would be true if Brown is guilty. So Smith's both statements are true. But if Smith is all-truthful, that's okay. Then Jones is all-lying. So his statements are both false. "Brown didn't do it" is a lie, so Brown did do it. "Smith did it" is a lie, so Smith didn't. That's consistent. And Brown's statements: "I didn't do it" is lie, "Jones didn't do it" is true. So Brown is mixed. That seems okay.But wait, the problem says "it was later found out that one of them told the truth twice, another lied twice, and the third told the truth once and lied once." So in this scenario, Smith (truthful twice), Jones (lied twice), Brown (truth once, lie once). That matches. So Brown is the criminal.But hold on, let me check if there's any other possibilities. Let me think again.Is there a scenario where Smith is the criminal? Earlier, when I tried Smith as criminal, Brown and Jones both ended up truthful. Wait, if Smith is the criminal, then:- Smith's statements: "I didn't do it" (lie), "Brown did it" (lie). So Smith is lying twice.- Brown's statements: "I didn't do it" (truth), "Jones didn't do it" (truth, since Smith is the criminal). So Brown is truthful twice.- Jones's statements: "Brown didn't do it" (truth), "Smith did it" (truth). So Jones is also truthful twice.But that can't be, because two people can't be truthful twice. So that's impossible. Therefore, Smith cannot be the criminal.If Jones is the criminal:- Jones's statements: "Brown didn't do it" (truth), "Smith did it" (lie). So Jones is mixed.- Brown's statements: "I didn't do it" (truth), "Jones didn't do it" (lie). So Brown is mixed.- Smith's statements: "I didn't do it" (truth), "Brown did it" (lie). So Smith is mixed.Three mixed people. Not allowed.Therefore, the only possible solution is Brown being the criminal.Wait, but let's check if there is another possibility where someone else could be the criminal with different truth-teller assignments.Alternatively, maybe the all-truthful person is someone else.Wait, if Jones is the criminal, then:- Jones's statements: 1. "Brown didn't do it" → true. 2. "Smith did it" → false. So Jones is mixed.But Brown's statements: "I didn't do it" → true. "Jones didn't do it" → false. So Brown is mixed.Smith's statements: "I didn't do it" → true. "Brown did it" → false. So Smith is mixed. All three mixed. Not allowed.If Smith is the criminal, as before, two truthful and one liar. Not allowed.If Brown is the criminal, one truthful (Smith), one liar (Jones), and Brown mixed. That works.Alternatively, maybe the all-truthful person is Jones? Let's see.Suppose someone else is all-truthful. Suppose Jones is all-truthful. Then both his statements are true: "Brown didn't do it" and "Smith did it." So Smith would be the criminal. But if Smith is the criminal, let's check:- Jones's statements: Both true. So Smith is criminal.- Smith's statements: "I didn't do it" → lie. "Brown did it" → lie. So Smith is all-lying.But then Brown's statements: "I didn't do it" → true. "Jones didn't do it" → true (since Smith is criminal). So Brown would be all-truthful. But Jones was supposed to be all-truthful, so that's two people all-truthful. Conflict. Therefore, impossible.Alternatively, if Jones is all-truthful, then Smith is criminal. But then Brown would also be all-truthful. Not allowed.Alternatively, if Smith is all-truthful, then his statements "I didn't do it" and "Brown did it" are both true. Therefore, Smith is innocent and Brown is guilty. So Brown is the criminal. Then:- Smith is all-truthful.- Brown's statements: "I didn't do it" → lie. "Jones didn't do it" → true. So Brown is mixed.- Jones's statements: "Brown didn't do it" → lie. "Smith did it" → lie. So Jones is all-lying.Which matches the initial case. So that works.Alternatively, if Brown is all-truthful, then both his statements are true: "I didn't do it" and "Jones didn't do it." Therefore, Brown and Jones are innocent, so Smith is the criminal. Then:- Brown is all-truthful.- Smith's statements: "I didn't do it" → lie. "Brown did it" → lie. So Smith is all-lying.- Jones's statements: "Brown didn't do it" → true. "Smith did it" → true. So Jones is all-truthful. But this would mean two all-truthfuls (Brown and Jones), which is impossible.Therefore, the only valid scenario is when Smith is all-truthful, Jones is all-lying, and Brown is mixed, leading to Brown being the criminal.Therefore, the answer should be Brown.Final AnswerThe criminal is boxed{Brown}.</think>To determine who committed the crime among Brown, Jones, and Smith, we analyze their statements and the given conditions: one told the truth twice, another lied twice, and the third told the truth once and lied once.Statements:- Brown: 1. "I didn't do it." 2. "Jones didn't do it."- Smith: 1. "I didn't do it." 2. "Brown did it."- Jones: 1. "Brown didn't do it." 2. "Smith did it."Case Analysis:1. Assuming Brown is the criminal: - Brown's statements: 1. "I didn't do it." (Lie) 2. "Jones didn't do it." (Truth) - Brown is mixed (1 lie, 1 truth). - Smith's statements: 1. "I didn't do it." (Truth) 2. "Brown did it." (Truth) - Smith is all-truthful. - Jones's statements: 1. "Brown didn't do it." (Lie) 2. "Smith did it." (Lie) - Jones is all-lying. - This satisfies the conditions: one all-truthful (Smith), one all-lying (Jones), and one mixed (Brown).2. Assuming Smith is the criminal: - Smith's statements: 1. "I didn't do it." (Lie) 2. "Brown did it." (Lie) - Smith is all-lying. - Brown's statements: 1. "I didn't do it." (Truth) 2. "Jones didn't do it." (Truth) - Brown is all-truthful. - Jones's statements: 1. "Brown didn't do it." (Truth) 2. "Smith did it." (Truth) - Jones is all-truthful. - This results in two all-truthfuls (Brown and Jones), which is invalid.3. Assuming Jones is the criminal: - Jones's statements: 1. "Brown didn't do it." (Truth) 2. "Smith did it." (Lie) - Jones is mixed. - Brown's statements: 1. "I didn't do it." (Truth) 2. "Jones didn't do it." (Lie) - Brown is mixed. - Smith's statements: 1. "I didn't do it." (Truth) 2. "Brown did it." (Lie) - Smith is mixed. - This results in three mixed, which is invalid.The only valid scenario is when Brown is the criminal, leading to Smith being all-truthful, Jones being all-lying, and Brown being mixed.boxed{Brown},<think>Okay, let's try to figure out who committed the crime among Brown, Jones, and Smith. The problem states that each of them made two statements, and we know that one person told the truth twice, another lied twice, and the third told the truth once and lied once. So, our job is to determine who is the criminal based on their statements and the given conditions about their truth-telling.First, let's list out all the statements each person made:- Brown said: 1. "I didn't do it." 2. "Jones didn't do it."- Smith said: 1. "I didn't do it." 2. "Brown did it."- Jones said: 1. "Brown didn't do it." 2. "Smith did it."Now, the key here is that each person's two statements must be evaluated for their truthfulness, and we have to match the conditions: one person is all truthful, one is all lies, and the third is a mix.Since the criminal is one of them, only one person is guilty. So, first, let's assume each one is guilty in turn and see if the statements fit the truth-telling conditions.Case 1: Suppose Brown is the criminal.If Brown did it, then let's check each person's statements.- Brown's statements: 1. "I didn't do it." – This is a lie. 2. "Jones didn't do it." – If Brown is the criminal, then Jones is innocent, so this is a truth. So Brown lied once and told the truth once. That would make him the one who told the truth once and lied once.- Smith's statements: 1. "I didn't do it." – True, because we're assuming Brown is guilty. 2. "Brown did it." – True. So Smith told two truths. But we already have Brown as the truth/lie mix. Then Smith would be the one who told the truth twice, which is possible.- Jones's statements: 1. "Brown didn't do it." – Lie, because Brown did it. 2. "Smith did it." – Lie, because Smith didn't do it. So Jones told two lies. That fits the remaining condition: someone lied twice. So in this scenario: - Smith: 2 truths - Jones: 2 lies - Brown: 1 truth, 1 lieThis satisfies all the given conditions. However, let's check other possibilities to make sure.Case 2: Suppose Smith is the criminal.If Smith did it, then:- Smith's statements: 1. "I didn't do it." – Lie 2. "Brown did it." – Lie (since Smith did it). So Smith would be the one who lied twice.- Brown's statements: 1. "I didn't do it." – True (Smith is guilty) 2. "Jones didn't do it." – True (Smith is guilty). So Brown told two truths. Then Brown is the truthful one.- Jones's statements: 1. "Brown didn't do it." – True (since Smith is guilty) 2. "Smith did it." – True (Smith is guilty). So Jones also told two truths. But we can't have two people telling the truth twice. So this contradicts the given conditions. Therefore, Smith cannot be the criminal.Case 3: Suppose Jones is the criminal.If Jones did it, then:- Jones's statements: 1. "Brown didn't do it." – True (Jones is guilty) 2. "Smith did it." – Lie (Jones did it). So Jones told one truth and one lie.- Brown's statements: 1. "I didn't do it." – True (Jones is guilty) 2. "Jones didn't do it." – Lie (Jones is guilty). So Brown told one truth and one lie.- Smith's statements: 1. "I didn't do it." – True (Jones is guilty) 2. "Brown did it." – Lie (Jones is guilty). So Smith also told one truth and one lie.But according to the problem, we need one person with two truths, one with two lies, and one with one of each. Here, all three would have one truth and one lie, which doesn't fit. Therefore, Jones can't be the criminal.So, from these three cases, only when Brown is the criminal do we satisfy the conditions:- Smith tells two truths.- Jones tells two lies.- Brown tells one truth and one lie.Wait a minute, but hold on. Let me double-check the first case again.If Brown is the criminal:- Brown's first statement "I didn't do it" is a lie.- Brown's second statement "Jones didn't do it" is a truth (since Brown is the criminal). So Brown has 1 lie, 1 truth.- Smith's first statement "I didn't do it" is true (because Brown did it).- Smith's second statement "Brown did it" is true. So Smith has two truths.- Jones's first statement "Brown didn't do it" is a lie.- Jones's second statement "Smith did it" is a lie (since Brown did it). So Jones has two lies.So yes, that works. Each of them falls into a different category: Smith (2 truths), Jones (2 lies), Brown (1 truth, 1 lie). So the criminal is Brown.But wait, hold on! Let me check again. If Brown is the criminal, then Smith's second statement "Brown did it" is true. So Smith is telling the truth twice, which is allowed. Jones's two statements are both lies, which is allowed. Brown's statements are one lie and one truth, which is the remaining category.But let me think again if there's another possibility. For example, if someone else is the criminal, but maybe the categories shift. Wait, in case 1, we saw that if Brown is guilty, the conditions are satisfied. In case 2, Smith being guilty leads to a contradiction because both Brown and Jones would have to tell two truths. In case 3, Jones being guilty causes everyone to have 1 truth and 1 lie.Therefore, the only viable solution is Brown being the criminal. Wait, but let's check the original problem again. The answer might be Smith or Jones, but according to my analysis, Brown is the culprit. Hmm.Wait, but in the case where Brown is the criminal, Jones is lying on both statements. So Jones's first statement is "Brown didn't do it" which is a lie, and "Smith did it," which is also a lie. That's correct.Smith is truthful on both: "I didn't do it" and "Brown did it." Correct.Brown's statements: "I didn't do it" (lie) and "Jones didn't do it" (truth). So he's the one with one lie and one truth.So that all checks out. Therefore, Brown is the criminal.Wait, but wait a second. Let me check the logic again. If Brown is the criminal, then Jones is innocent, so Brown's second statement "Jones didn't do it" is true. So Brown's two statements: lie and truth. Smith's statements: both truths. Jones's statements: both lies. That works.But hold on. Wait, when Jones says "Smith did it," which is a lie if Brown did it. So that's correct.But let me consider another angle. Suppose we consider that perhaps the categories of truth-tellers aren't assigned to the same person as the criminal. For example, maybe even if someone is the criminal, they can still tell the truth in their statements. But in reality, the criminal would have to lie if they claim they didn't do it. Wait, but the criminal could be lying or telling the truth in their statements. Wait, but if the criminal is Brown, and he says "I didn't do it," that's a lie, which is expected. But Smith, an innocent person, says "I didn't do it," which is true. Final Answerboxed{25}]

🔑:Think about whether Jones could have been the thief.## SolutionThe thief is Brown. If we consider the three possibilities - "Brown stole it", "Smith stole it", and "Jones stole it" - we see that only in the case of "the thief is Brown" do the conditions of the problem hold. If the thief is Smith, then both Brown and Jones told the truth. If the thief is Jones, then both Brown and Smith simultaneously told the truth and lied.

🧠:[<think>Okay, so I have this problem here where I need to find the value of 'a'. The given equation is (7m² -14m +a)(3n² -6n -7) =8, and m is 1+√2, n is 1-√2. Hmm, let's see. First, maybe I should compute the values of m and n, but since they are given as 1+√2 and 1-√2, maybe there's a relationship between them. Oh, right! They are conjugates. That might come in handy.Let me write down m and n:m = 1 + √2n = 1 - √2So, m + n = (1 + √2) + (1 - √2) = 2And m * n = (1 + √2)(1 - √2) = 1 - (√2)² = 1 - 2 = -1That's useful. Maybe I can use these to simplify m² and n² terms without expanding everything.Looking at the equation: (7m² -14m +a)(3n² -6n -7)=8Let me first look at each bracket separately. Maybe I can simplify each part first.Starting with the first bracket: 7m² -14m +aFactor out a 7: 7(m² - 2m) + aHmm, m² - 2m. Wait, m = 1 + √2. Let's compute m - 1 = √2. So, maybe m² - 2m can be expressed in terms of (m -1)²?Let me check:(m -1)² = (√2)² = 2But expanding (m -1)² = m² - 2m +1So, m² -2m = (m -1)² -1 = 2 -1 =1Wait, that's interesting. So m² -2m =1. Therefore, 7(m² -2m) +a =7*1 +a =7 +aWait, so the first bracket simplifies to 7 + a? That seems too simple, but let me verify.Given m =1 +√2,Compute m²:m² = (1 +√2)² =1 + 2√2 +2 =3 + 2√2Then m² -2m = (3 + 2√2) -2*(1 +√2) =3 +2√2 -2 -2√2 =1Yes, exactly. So m² -2m =1, so 7(m² -2m) =7*1=7. Therefore, 7m² -14m +a =7 +a. Cool, that simplifies the first bracket to (7 +a).Now the second bracket: 3n² -6n -7Let me try a similar approach. Let's factor out 3: 3(n² - 2n) -7n =1 -√2, so n -1 = -√2Compute n² -2n:n² -2n = (n -1)² -1(n -1)² = (-√2)² =2Therefore, n² -2n =2 -1=1Therefore, 3(n² -2n) -7 =3*1 -7=3 -7= -4Wait, so the second bracket simplifies to -4. Let me verify that.Compute n²: (1 -√2)² =1 -2√2 +2=3 -2√2Then n² -2n = (3 -2√2) -2*(1 -√2)=3 -2√2 -2 +2√2=1Yes, so 3*(1)=3, then subtract 7 gives -4. Correct.Therefore, the original equation becomes:(7 +a)*(-4) =8So, -4*(7 +a)=8Divide both sides by -4:7 +a =8 / (-4)= -2Then subtract 7:a = -2 -7 = -9Wait, so a is -9?But let me double-check all steps to make sure I didn't make any mistakes.First, simplifying 7m² -14m +a:We found m² -2m =1, so 7*(1)=7, so 7 +a. Correct.Second bracket: 3n² -6n -7: n² -2n=1, so 3*1=3, 3 -7=-4. Correct.Multiply: (7 +a)*(-4)=8So (7 +a)=8/(-4)= -2Therefore, a= -2 -7= -9Seems correct. Let me check with the actual values of m and n to confirm.First, compute m=1 +√2≈1 +1.414≈2.414Compute m²≈5.8287m²≈40.79614m≈14*2.414≈33.796So 7m² -14m ≈40.796 -33.796≈7Then 7m² -14m +a≈7 +a. So that holds numerically.Similarly, compute n=1 -√2≈1 -1.414≈-0.414n²≈0.1713n²≈0.5146n≈-2.485So 3n² -6n≈0.514 -(-2.485)=0.514 +2.485≈3.0Then subtract 7: 3.0 -7≈-4. Correct.Multiply (7 +a)*(-4)=8If a=-9, then 7 +a= -2. Then (-2)*(-4)=8. Correct.Yes, that works. So the answer is a= -9.Final AnswerThe value of ( a ) is boxed{-9}.</think>Given ( m = 1 + sqrt{2} ) and ( n = 1 - sqrt{2} ), we need to find the value of ( a ) such that (left(7m^2 - 14m + aright)left(3n^2 - 6n - 7right) = 8).First, we simplify each bracket separately.For the first bracket ( 7m^2 - 14m + a ):- Note that ( m = 1 + sqrt{2} ).- Compute ( m - 1 = sqrt{2} ).- Then, ( (m - 1)^2 = (sqrt{2})^2 = 2 ).- Expanding ( (m - 1)^2 ), we get ( m^2 - 2m + 1 = 2 ).- Therefore, ( m^2 - 2m = 1 ).- Substitute this into the expression: ( 7(m^2 - 2m) + a = 7 cdot 1 + a = 7 + a ).For the second bracket ( 3n^2 - 6n - 7 ):- Note that ( n = 1 - sqrt{2} ).- Compute ( n - 1 = -sqrt{2} ).- Then, ( (n - 1)^2 = (-sqrt{2})^2 = 2 ).- Expanding ( (n - 1)^2 ), we get ( n^2 - 2n + 1 = 2 ).- Therefore, ( n^2 - 2n = 1 ).- Substitute this into the expression: ( 3(n^2 - 2n) - 7 = 3 cdot 1 - 7 = 3 - 7 = -4 ).Now, substituting the simplified forms into the given equation:[(7 + a)(-4) = 8]Solving for ( a ):[-4(7 + a) = 8 7 + a = frac{8}{-4} 7 + a = -2 a = -2 - 7 a = -9]Thus, the value of ( a ) is (boxed{-9}).,<think>Okay, so I need to find the value of 'a' such that when I multiply these two expressions, (7m² -14m + a) and (3n² -6n -7), the product is 8. The given values for m and n are m = 1 + √2 and n = 1 - √2. Hmm, let me think step by step.First, maybe I should compute m and n. Wait, m is 1 plus root 2, which is approximately 2.414, and n is 1 minus root 2, which is approximately -0.414. But maybe exact values are needed here. Since m and n are conjugates, perhaps there's a relationship between them that can simplify the problem?Let me note that m and n are roots of the quadratic equation x² - 2x -1 = 0. Because if we consider (x - (1 + √2))(x - (1 - √2)) = (x -1 -√2)(x -1 +√2) = (x -1)² - (√2)² = x² - 2x +1 -2 = x² -2x -1. So yes, m and n satisfy x² = 2x +1. That might be helpful for simplifying higher powers.So, for example, m² = 2m +1 and similarly n² = 2n +1. That could help reduce m² and n² in the expressions. Let me check that.Given m = 1 + √2. Then m² = (1 + √2)² = 1 + 2√2 + 2 = 3 + 2√2. On the other hand, 2m +1 = 2*(1 + √2) +1 = 2 + 2√2 +1 = 3 + 2√2. Yep, that's correct. Similarly for n: n =1 - √2. Then n² = (1 - √2)² = 1 - 2√2 + 2 = 3 - 2√2. And 2n +1 = 2*(1 - √2) +1 = 2 - 2√2 +1 = 3 - 2√2. So that works too. Good.Therefore, we can replace m² with 2m +1 and n² with 2n +1. That should simplify the expressions. Let me try that.Starting with the first expression: 7m² -14m + a. Replace m² with 2m +1:7*(2m +1) -14m + a = 14m +7 -14m +a = (14m -14m) +7 +a = 0 +7 +a = 7 + a. Oh, wow, that simplifies nicely! So the first expression simplifies to (7 + a). That's much easier.Now the second expression: 3n² -6n -7. Replace n² with 2n +1:3*(2n +1) -6n -7 = 6n +3 -6n -7 = (6n -6n) +3 -7 = 0 -4 = -4. So the second expression simplifies to -4. That's really helpful!So now, the original equation (7m² -14m +a)(3n² -6n -7) =8 simplifies to (7 + a)*(-4) =8. Therefore:(7 + a)*(-4) =8.Let me solve for a. Divide both sides by -4:7 + a = 8 / (-4) = -2.Therefore, a = -2 -7 = -9.Wait, that seems straightforward. Let me verify.First expression: 7m² -14m +a. If m² = 2m +1, then 7*(2m +1) =14m +7. Then subtract 14m: 14m +7 -14m =7. Then add a: 7 + a. Correct.Second expression: 3n² -6n -7. Since n² =2n +1, 3*(2n +1) =6n +3. Then subtract 6n: 6n +3 -6n =3. Then subtract 7: 3 -7 = -4. Correct.So then (7 + a)*(-4)=8. So 7 + a = -2. Thus a= -9. Seems correct. Let me double-check by plugging in the original values.Let's compute 7m² -14m +a when a=-9. m =1 +√2.Compute m²: (1 +√2)^2 =1 +2√2 +2=3 +2√2.Then 7m²=7*(3 +2√2)=21 +14√2.14m=14*(1 +√2)=14 +14√2.So 7m² -14m =21 +14√2 -14 -14√2= (21-14)+(14√2 -14√2)=7 +0=7.Then 7m² -14m +a =7 + (-9)= -2.Now compute 3n² -6n -7. n=1 -√2.n²=(1 -√2)^2=1 -2√2 +2=3 -2√2.3n²=3*(3 -2√2)=9 -6√2.6n=6*(1 -√2)=6 -6√2.So 3n² -6n=9 -6√2 -6 +6√2= (9-6)+(-6√2 +6√2)=3 +0=3.Then 3n² -6n -7=3 -7= -4.Multiply (-2)*(-4)=8, which matches the given equation. Therefore, a= -9 is correct.So the answer is boxed{-9}.Final AnswerThe value of ( a ) is boxed{-9}.</think>Given ( m = 1 + sqrt{2} ) and ( n = 1 - sqrt{2} ), we need to find the value of ( a ) such that (left(7m^2 - 14m + aright)left(3n^2 - 6n - 7right) = 8).First, we note that ( m ) and ( n ) are roots of the quadratic equation ( x^2 - 2x - 1 = 0 ). Therefore, we can use the relationships ( m^2 = 2m + 1 ) and ( n^2 = 2n + 1 ).Simplifying the expressions:1. For ( 7m^2 - 14m + a ): [ 7m^2 - 14m + a = 7(2m + 1) - 14m + a = 14m + 7 - 14m + a = 7 + a ]2. For ( 3n^2 - 6n - 7 ): [ 3n^2 - 6n - 7 = 3(2n + 1) - 6n - 7 = 6n + 3 - 6n - 7 = -4 ]Substituting these simplified expressions into the given equation:[(7 + a)(-4) = 8]Solving for ( a ):[7 + a = frac{8}{-4} = -2 implies a = -2 - 7 = -9]Thus, the value of ( a ) is (boxed{-9}).]

🔑:answer: -9

🧠:[<think>Okay, let's see. I need to simplify this big fraction:[frac{2005^{2}left(2004^{2}-2003right)}{left(2004^{2}-1right)left(2004^{3}+1right)} times frac{2003^{2}left(2004^{2}+2005right)}{2004^{3}-1}]Hmm, this looks complicated, but maybe there are ways to factor some parts here. Let me take it step by step.First, let me note that all the numbers are close to 2004 or 2005 or 2003. Maybe substituting variables could help? Let me try setting x = 2004. Then:- 2003 = x - 1- 2005 = x + 1So substituting these into the expression, it becomes:Numerator first part: (x + 1)^2 * (x^2 - (x - 1))Denominator first fraction: (x^2 - 1) * (x^3 + 1)Second fraction numerator: (x - 1)^2 * (x^2 + (x + 1))Denominator second fraction: (x^3 - 1)So overall, the entire expression becomes:[frac{(x + 1)^2 cdot (x^2 - x + 1)}{(x^2 - 1)(x^3 + 1)} times frac{(x - 1)^2 cdot (x^2 + x + 1)}{(x^3 - 1)}]Now, let's simplify each part. Let me first compute each component:First numerator part: (x + 1)^2 * (x² - x + 1)First denominator part: (x² - 1)(x³ + 1)Second numerator part: (x - 1)^2 * (x² + x + 1)Second denominator part: (x³ - 1)Now, note that x² - 1 factors as (x - 1)(x + 1). Similarly, x³ + 1 and x³ - 1 can be factored using sum and difference of cubes:x³ + 1 = (x + 1)(x² - x + 1)x³ - 1 = (x - 1)(x² + x + 1)So substituting these factorizations into the denominators:First denominator becomes (x - 1)(x + 1) * (x + 1)(x² - x + 1) = (x - 1)(x + 1)^2(x² - x + 1)Second denominator becomes (x - 1)(x² + x + 1)So now, let's rewrite the entire expression with these factorizations:First fraction numerator: (x + 1)^2 * (x² - x + 1)First fraction denominator: (x - 1)(x + 1)^2(x² - x + 1)Second fraction numerator: (x - 1)^2 * (x² + x + 1)Second fraction denominator: (x - 1)(x² + x + 1)So putting it all together:First fraction simplifies by canceling (x + 1)^2 * (x² - x + 1) in numerator and denominator, leaving 1/(x - 1)Similarly, the second fraction simplifies by canceling (x - 1)(x² + x + 1) in denominator with numerator's (x - 1)^2 * (x² + x + 1), leaving (x - 1) in the numerator.Therefore, the entire expression becomes:(1/(x - 1)) * (x - 1) = 1Wait, so does this all simplify to 1? Let me check again.First fraction: [ (x + 1)^2 (x² - x + 1) ] / [ (x² - 1)(x³ + 1) ]But (x² - 1) is (x - 1)(x + 1), and (x³ + 1) is (x + 1)(x² - x + 1). So denominator is (x - 1)(x + 1)*(x + 1)(x² - x + 1) = (x - 1)(x + 1)^2(x² - x + 1). The numerator is (x + 1)^2(x² - x + 1). So dividing, numerator and denominator cancel except for 1/(x - 1).Second fraction: [ (x - 1)^2(x² + x + 1) ] / (x³ - 1)But x³ - 1 is (x - 1)(x² + x + 1), so denominator is (x - 1)(x² + x + 1). Numerator is (x - 1)^2(x² + x + 1). So dividing, we get (x - 1) left.Multiply the two simplified parts: [1/(x - 1)] * [x - 1] = 1.Therefore, the entire expression simplifies to 1. So substituting back x = 2004, the original expression simplifies to 1? That seems surprising but maybe correct.Let me verify with actual numbers. Let me try replacing x with a smaller number, like x = 2, to see if the expression holds.So if x = 2, then 2003 = 1, 2004 = 2, 2005 = 3.Original expression becomes:[3²*(2² -1)] / [(2² -1)(2³ +1)] * [1²*(2² +3)] / [2³ -1]Compute each part:First numerator: 9*(4 -1) = 9*3 = 27First denominator: (4 -1)(8 +1) = 3*9 = 27Second numerator: 1*(4 +3) = 7Second denominator: 8 -1 =7So expression becomes (27/27)*(7/7) = 1*1 =1. So indeed, with x=2, the result is 1. Hence, likely correct.Therefore, the original expression simplifies to 1.Final Answerboxed{1}</think>To simplify the given expression:[frac{2005^{2}left(2004^{2}-2003right)}{left(2004^{2}-1right)left(2004^{3}+1right)} times frac{2003^{2}left(2004^{2}+2005right)}{2004^{3}-1}]we substitute ( x = 2004 ). Then we have:- ( 2003 = x - 1 )- ( 2005 = x + 1 )Substituting these into the expression, we get:[frac{(x + 1)^2 cdot (x^2 - (x - 1))}{(x^2 - 1)(x^3 + 1)} times frac{(x - 1)^2 cdot (x^2 + (x + 1))}{(x^3 - 1)}]Simplifying each part:- The numerator of the first fraction becomes ( (x + 1)^2 cdot (x^2 - x + 1) )- The denominator of the first fraction becomes ( (x^2 - 1)(x^3 + 1) )- The numerator of the second fraction becomes ( (x - 1)^2 cdot (x^2 + x + 1) )- The denominator of the second fraction becomes ( x^3 - 1 )Factoring the denominators:- ( x^2 - 1 = (x - 1)(x + 1) )- ( x^3 + 1 = (x + 1)(x^2 - x + 1) )- ( x^3 - 1 = (x - 1)(x^2 + x + 1) )Substituting these factorizations into the expression, we get:- The first fraction simplifies to ( frac{(x + 1)^2 cdot (x^2 - x + 1)}{(x - 1)(x + 1)^2 cdot (x^2 - x + 1)} = frac{1}{x - 1} )- The second fraction simplifies to ( frac{(x - 1)^2 cdot (x^2 + x + 1)}{(x - 1)(x^2 + x + 1)} = x - 1 )Multiplying the simplified fractions:[frac{1}{x - 1} times (x - 1) = 1]Thus, the original expression simplifies to:[boxed{1}],<think>Okay, let's see. I have this expression to simplify:[frac{2005^{2}left(2004^{2}-2003right)}{left(2004^{2}-1right)left(2004^{3}+1right)} times frac{2003^{2}left(2004^{2}+2005right)}{2004^{3}-1}]Hmm, that's a bit complex. Let me break it down step by step. Maybe I can factor some parts of it to see if there's cancellation or simplification.First, let's look at the denominators and numerators to see if they can be factored. Starting with the first fraction's denominator: (2004² - 1)(2004³ + 1). I recognize that 2004² - 1 is a difference of squares, so it can be factored as (2004 - 1)(2004 + 1) = 2003 * 2005. That's helpful.Then, the second part of the denominator is 2004³ + 1. That's a sum of cubes, right? The formula for a³ + b³ is (a + b)(a² - ab + b²). So here, a = 2004, b = 1, so it factors into (2004 + 1)(2004² - 2004*1 + 1²) = 2005*(2004² - 2004 + 1). Let me calculate that inside part: 2004² - 2004 + 1. Not sure if that's useful yet, but maybe.Similarly, the second denominator in the second fraction is 2004³ - 1. That's a difference of cubes. The formula is a³ - b³ = (a - b)(a² + ab + b²). So here, a = 2004, b = 1, so that factors into (2004 - 1)(2004² + 2004*1 + 1²) = 2003*(2004² + 2004 + 1).So, let's note these factorizations:First denominator: (2004² - 1)(2004³ + 1) = 2003*2005*(2005*(2004² - 2004 + 1)) = 2003*2005²*(2004² - 2004 + 1)Second denominator: 2004³ - 1 = 2003*(2004² + 2004 + 1)Now, the numerators:First numerator: 2005²*(2004² - 2003)Second numerator: 2003²*(2004² + 2005)So putting everything together, the entire expression becomes:[frac{2005^2 times (2004^2 - 2003)}{2003 times 2005^2 times (2004^2 - 2004 + 1)} times frac{2003^2 times (2004^2 + 2005)}{2003 times (2004^2 + 2004 + 1)}]Wait, let me check that again. The first denominator factors into 2003*2005*(2004³ + 1) which is 2003*2005*2005*(2004² - 2004 + 1). So that's 2003*2005²*(2004² - 2004 + 1). Correct.Then the second denominator is 2003*(2004² + 2004 + 1). So the denominators combined are 2003*2005²*(2004² - 2004 + 1)*2003*(2004² + 2004 + 1). Wait, but actually, in the original expression, the denominators are multiplied together, right? So (2004² - 1)(2004³ + 1) multiplied by (2004³ - 1). Wait, no, actually the expression is:First fraction: numerator / denominator, then multiplied by second fraction: numerator / denominator.So overall, the entire expression's numerator is 2005²*(2004² - 2003)*2003²*(2004² + 2005), and the denominator is (2003*2005²*(2004² - 2004 + 1)) * (2003*(2004² + 2004 + 1))).So let me write that out:Numerator: 2005² * (2004² - 2003) * 2003² * (2004² + 2005)Denominator: 2003 * 2005² * (2004² - 2004 + 1) * 2003 * (2004² + 2004 + 1)Simplify numerator and denominator:Numerator: 2003² * 2005² * (2004² - 2003) * (2004² + 2005)Denominator: 2003² * 2005² * (2004² - 2004 + 1) * (2004² + 2004 + 1)So the 2003² and 2005² terms cancel out from numerator and denominator. So we're left with:[frac{(2004² - 2003)(2004² + 2005)}{(2004² - 2004 + 1)(2004² + 2004 + 1)}]Okay, now the key is to simplify this remaining fraction. Let's look at the numerator and the denominator.First, let's compute the terms in the numerator:(2004² - 2003) and (2004² + 2005). Let's see if these can be related to the denominator terms.Denominator terms are (2004² - 2004 + 1) and (2004² + 2004 + 1). Hmm.Wait, let's compute the numerator first:2004² - 2003 = 2004² - 2004 + 1 (since -2003 = -2004 +1). Because 2004² - 2004 +1 = 2004² -2004 +1, which is exactly the first denominator term.Similarly, 2004² + 2005 = 2004² + 2004 +1 (since 2005 = 2004 +1). So 2004² + 2004 +1 is the second denominator term.So that means:Numerator: (2004² - 2004 +1)(2004² + 2004 +1)Denominator: (2004² - 2004 +1)(2004² + 2004 +1)Therefore, the entire fraction simplifies to 1.Wait, hold on. Let me check that again.If 2004² - 2003 = 2004² - 2004 +1, because 2004² -2003 = 2004² - (2004 -1) = 2004² -2004 +1. Yes, that's correct.Similarly, 2004² + 2005 = 2004² + (2004 +1) = 2004² +2004 +1. Correct.Therefore, the numerator becomes (2004² -2004 +1)(2004² +2004 +1) which is exactly the denominator. Therefore, the fraction is 1.So after all the cancellation, the entire expression simplifies to 1.Wait, but let me double-check to make sure I didn't make any mistake in the factoring.Original numerator terms:2004² - 2003: Let's compute 2004² -2003. If 2004² = (2004)(2004). Let's compute 2004*2004:Well, 2004*2004 = (2000 + 4)^2 = 2000² + 2*2000*4 +4² = 4,000,000 + 16,000 + 16 = 4,016,016.So 2004² = 4,016,016.Then 2004² -2003 = 4,016,016 -2003 = 4,014,013.Now, 2004² -2004 +1: 4,016,016 -2004 +1 = 4,016,016 -2004 is 4,014,012, then +1 is 4,014,013. So that matches. So yes, 2004² -2003 = 2004² -2004 +1.Similarly, 2004² +2005: 4,016,016 +2005 = 4,018,021.2004² +2004 +1: 4,016,016 +2004 +1 = 4,018,021. So that also matches. So indeed, numerator factors into the same as denominator. Therefore, the fraction is 1.Therefore, after all cancellations, the entire original expression simplifies to 1.So the answer is 1, boxed.Final Answerboxed{1}</think>To simplify the given expression:[frac{2005^{2}left(2004^{2}-2003right)}{left(2004^{2}-1right)left(2004^{3}+1right)} times frac{2003^{2}left(2004^{2}+2005right)}{2004^{3}-1}]we start by factoring the denominators and numerators.1. Factor the denominators: - (2004^2 - 1) is a difference of squares: ( (2004 - 1)(2004 + 1) = 2003 cdot 2005 ) - (2004^3 + 1) is a sum of cubes: ( (2004 + 1)(2004^2 - 2004 cdot 1 + 1^2) = 2005(2004^2 - 2004 + 1) ) - (2004^3 - 1) is a difference of cubes: ( (2004 - 1)(2004^2 + 2004 cdot 1 + 1^2) = 2003(2004^2 + 2004 + 1) )2. Factor the numerators: - (2004^2 - 2003) can be rewritten as (2004^2 - 2004 + 1) - (2004^2 + 2005) can be rewritten as (2004^2 + 2004 + 1)Substituting these factorizations into the expression, we get:[frac{2005^2 cdot (2004^2 - 2004 + 1)}{2003 cdot 2005^2 cdot (2004^2 - 2004 + 1)} times frac{2003^2 cdot (2004^2 + 2004 + 1)}{2003 cdot (2004^2 + 2004 + 1)}]Simplifying the numerators and denominators, we cancel out the common factors:- (2003^2) and (2005^2) in the numerator and denominator cancel out.- The terms ((2004^2 - 2004 + 1)) and ((2004^2 + 2004 + 1)) in the numerator and denominator also cancel out.This leaves us with:[frac{(2004^2 - 2004 + 1)(2004^2 + 2004 + 1)}{(2004^2 - 2004 + 1)(2004^2 + 2004 + 1)} = 1]Thus, the simplified expression is:[boxed{1}]]

🔑:15. Ans: 1Let t=2004. Then the expression is equal tobegin{aligned}& frac{(t+1)^{2}left(t^{2}-t+1right)}{left(t^{2}-1right)left(t^{3}+1right)} times frac{(t-1)^{2}left(t^{2}+t+1right)}{t^{3}-1} = & frac{(t+1)^{2}left(t^{2}-t+1right)}{(t+1)(t-1)(t+1)left(t^{2}-t+1right)} times frac{(t-1)^{2}left(t^{2}+t+1right)}{(t-1)left(t^{2}+t+1right)}=1 .end{aligned}