🧠:[<think>Okay, let's see. I need to figure out the relationship between conditions A and B. The question is about whether a = b (option A) is sufficient, necessary, both, or neither for A = B (option B), where A and B are sets defined by quadratic inequalities. Hmm.First, let me recall that for a quadratic inequality like x² - x + c ≤ 0, the solution set depends on the discriminant of the quadratic equation x² - x + c = 0. The discriminant D is 1 - 4c. If D is positive, the quadratic has two real roots, and the inequality x² - x + c ≤ 0 will hold between the two roots. If D is zero, there's exactly one real root, and the inequality holds only at that point. If D is negative, there are no real roots, meaning the quadratic is always positive (since the coefficient of x² is positive), so the inequality x² - x + c ≤ 0 would have no solution.So, for the sets A and B to be non-empty, their respective quadratics must have real roots. That means the discriminants must be non-negative. For set A, discriminant D_A = 1 - 4a ≥ 0 ⇒ a ≤ 1/4. Similarly, for set B, discriminant D_B = 1 - 4b ≥ 0 ⇒ b ≤ 1/4.Now, if a and b are both greater than 1/4, then both A and B would be empty sets, right? Because the quadratic would never be less than or equal to zero. So in that case, even if a ≠ b, A and B would both be empty, hence equal. Hmm, interesting. So if a and b are both greater than 1/4, A and B are equal (both empty sets), but a doesn't have to equal b. That suggests that a = b is not necessary for A = B. Therefore, option A is not a necessary condition for option B. But if a = b, then certainly the sets A and B would be the same, right? Because they'd have the same quadratic inequality. So a = b is sufficient for A = B. But since there are cases where a ≠ b but A = B (like when both a and b are greater than 1/4), then a = b is not necessary. Therefore, option A is a sufficient condition but not necessary. So the answer should be option A: "Option A is a sufficient condition but not a necessary condition for option B."Wait, but let me check if there are other possibilities where a ≠ b but A = B. Suppose a and b are both less than or equal to 1/4. Then the quadratics have real roots. For set A, the roots are [1 ± sqrt(1 - 4a)] / 2, and similarly for set B. So for A and B to be equal, the intervals defined by their roots must be the same. So the roots must be the same. So the quadratics x² - x + a and x² - x + b must have the same roots. Since they have the same leading coefficient and the same linear term coefficient, if they have the same roots, then the constant terms must be the same. Therefore, a must equal b. Therefore, if a and b are both ≤ 1/4, then A = B implies a = b. However, if both a and b are >1/4, then A and B are both empty sets, so A = B regardless of a and b. Therefore, in that case, a ≠ b but A = B. Therefore, the only time A = B can happen with a ≠ b is when both a and b are >1/4.Therefore, if a = b, then A = B (sufficient). But A = B can also happen when a ≠ b, as long as both a and b are >1/4. Therefore, a = b is sufficient but not necessary. So the answer is A.Wait, but the options given are labeled A, B, C, D. Wait the options for the answer are labeled A, B, C, D. Let me check:Original options:A: Option A is a sufficient condition but not a necessary condition for option B.B: Option A is a necessary condition but not a sufficient condition for option B.C: Option A is both a necessary and a sufficient condition for option B.D: Option A is neither a sufficient nor a necessary condition for option B.So according to my reasoning, the answer is A. Let me confirm once more.If a = b, then regardless of the value of a, A and B will be the same set. So that's clear. So a = b is sufficient.But as for necessary, if a ≠ b, can A still equal B? Yes. For example, if a and b are both greater than 1/4, then A and B are both empty sets, so A = B even if a ≠ b. Therefore, a = b is not necessary. Therefore, A is sufficient but not necessary. So the answer is option A.But hold on, let me think of another case. Suppose a and b are different but both less than or equal to 1/4, but such that the roots of the quadratics are the same. Wait, but can two quadratics with the same leading coefficient and same linear coefficient but different constants have the same roots? No, because the roots are determined by the coefficients. If x² - x + a and x² - x + b have the same roots, then a must equal b. Because the roots depend on the discriminant. If the roots are the same, then discriminant and the roots must be same. The roots for x² - x + a = 0 are [1 ± sqrt(1 - 4a)] / 2. For these to be equal to [1 ± sqrt(1 - 4b)] / 2, we must have sqrt(1 - 4a) = sqrt(1 - 4b) ⇒ 1 - 4a = 1 - 4b ⇒ a = b. So if a and b are both ≤ 1/4, then A = B only if a = b. But if a and b are both >1/4, then A = B even if a ≠ b. Therefore, the cases where A = B are either a = b (and a, b ≤1/4), or a and b both >1/4 (regardless of their relation). Therefore, a = b is not necessary for A = B because you can have A = B when both a and b are >1/4. So yes, a = b is sufficient, but not necessary. Therefore, answer is option A.But wait, the answer options are labeled A, B, C, D. The answer according to the user's options is A: "Option A is a sufficient condition but not a necessary condition for option B." So that's correct. So the correct answer is A.Final Answerboxed{A}</think>Given the sets ( A = { x in mathbb{R} mid x^2 - x + a leq 0 } ) and ( B = { x in mathbb{R} mid x^2 - x + b leq 0 } ), we need to determine the relationship between the conditions ( a = b ) (Option A) and ( A = B ) (Option B).1. Quadratic Inequalities Analysis: - The quadratic ( x^2 - x + c leq 0 ) has real solutions if its discriminant is non-negative: ( 1 - 4c geq 0 ), i.e., ( c leq frac{1}{4} ). - If ( c > frac{1}{4} ), the quadratic has no real solutions, so the set is empty.2. Case Analysis: - Case 1: If ( a = b ), then the quadratic inequalities defining ( A ) and ( B ) are identical, hence ( A = B ). This shows ( a = b ) is a sufficient condition for ( A = B ). - Case 2: If ( a neq b ), ( A ) and ( B ) can still be equal if both ( a ) and ( b ) are greater than ( frac{1}{4} ). In this case, both sets ( A ) and ( B ) are empty sets, hence equal. This shows ( a = b ) is not a necessary condition for ( A = B ).3. Conclusion: - ( a = b ) is sufficient for ( A = B ) because identical quadratics yield the same solution set. - ( a = b ) is not necessary for ( A = B ) because different ( a ) and ( b ) can result in both sets being empty (when ( a, b > frac{1}{4} )).Thus, Option A is a sufficient condition but not a necessary condition for Option B.[boxed{A}],<think>Okay, so I need to figure out whether option A (a = b) is a sufficient, necessary, both, or neither condition for option B (A = B). The sets A and B are defined as the solutions to the quadratic inequalities x² - x + a ≤ 0 and x² - x + b ≤ 0, respectively. First, let me recall that a quadratic inequality like x² + px + q ≤ 0 will have a solution set that depends on the discriminant of the quadratic. The discriminant here is p² - 4q. If the discriminant is positive, the quadratic crosses the x-axis at two points, and the solution set is the interval between those two roots. If the discriminant is zero, there's exactly one real root, and the solution set is just that single point. If the discriminant is negative, the quadratic doesn't cross the x-axis at all, so the inequality x² + px + q ≤ 0 would have no solution if the quadratic opens upwards (since it's always positive) or all real numbers if it opens downwards. But in our case, the coefficient of x² is positive (1), so the parabola opens upwards. Therefore, if the discriminant is negative, the inequality x² - x + a ≤ 0 would have no solution. So, for the sets A and B to be non-empty, the discriminants of their quadratics must be non-negative. Let's compute those discriminants.For set A: discriminant D_A = (-1)² - 4*1*a = 1 - 4a. Similarly, for set B: D_B = 1 - 4b. For there to be real solutions, we need D_A ≥ 0 and D_B ≥ 0. So, 1 - 4a ≥ 0 ⇒ a ≤ 1/4, and similarly b ≤ 1/4. If a > 1/4, set A is empty, and same for set B if b > 1/4.Assuming that both a and b are ≤ 1/4, the sets A and B are intervals between the roots of their respective quadratics. Let's find those roots.For set A: the quadratic equation x² - x + a = 0. Using the quadratic formula, the roots are [1 ± sqrt(1 - 4a)] / 2. Similarly for set B: roots are [1 ± sqrt(1 - 4b)] / 2.Therefore, set A is the interval [ (1 - sqrt(1 - 4a))/2 , (1 + sqrt(1 - 4a))/2 ], and set B is [ (1 - sqrt(1 - 4b))/2 , (1 + sqrt(1 - 4b))/2 ].For these intervals to be equal, their endpoints must be equal. That is:(1 - sqrt(1 - 4a))/2 = (1 - sqrt(1 - 4b))/2and(1 + sqrt(1 - 4a))/2 = (1 + sqrt(1 - 4b))/2Subtracting the first equation from the second gives sqrt(1 - 4a) = sqrt(1 - 4b). Since sqrt is injective in non-negative numbers, squaring both sides gives 1 - 4a = 1 - 4b ⇒ a = b. Therefore, if A = B, then it must be that a = b. So, a = b is a necessary condition for A = B. But is it sufficient? Suppose a = b. Then, the intervals A and B would be defined by the same quadratic, so they would have the same roots and hence the same interval. Therefore, A = B. Thus, a = b is also a sufficient condition for A = B. Wait, but hold on. What if a or b is greater than 1/4? For example, suppose a = b = 1/2. Then, both discriminants would be 1 - 4*(1/2) = 1 - 2 = -1, which is negative. Therefore, both sets A and B would be empty. So, even though a = b, they both result in empty sets. In that case, A = B because both are empty. So, even if a = b and a > 1/4, A and B are both empty, hence equal. But wait, if a = b, regardless of their values, if they are equal, then A and B will be the same. If a = b ≤ 1/4, then they are intervals as above. If a = b > 1/4, then they are both empty. So, in either case, A = B if a = b. Therefore, a = b is sufficient for A = B.Conversely, if A = B, then either both are empty or both are intervals. If both are empty, then a and b must both be > 1/4. But they could be different as long as both are >1/4. Wait, no. Wait, if A = B, and they are both empty, then it's possible that a and b are different but both greater than 1/4. For example, let a = 1/2 and b = 3/4. Then, A is empty because 1 - 4*(1/2) = -1 < 0, and B is empty because 1 - 4*(3/4) = 1 - 3 = -2 < 0. So, A and B are both empty, hence equal, even though a ≠ b. Wait, that contradicts the previous conclusion. So, if a and b are different but both >1/4, then A and B are both empty, hence equal. Therefore, in that case, A = B even though a ≠ b. Therefore, the condition a = b is not necessary for A = B. But earlier, I thought that if A = B, then a must equal b, but that was under the assumption that the sets are non-empty. If the sets are empty, then A = B regardless of a and b, as long as both a and b make the discriminant negative. So, if a ≠ b but both >1/4, then A and B are both empty, hence equal. Therefore, A = B does not require a = b. Therefore, a = b is not a necessary condition for A = B. Wait, that's conflicting with my earlier conclusion. Let me clarify. Case 1: If both A and B are non-empty. Then, for them to be equal, their intervals must be equal, which requires a = b. Case 2: If both A and B are empty. Then, they are equal (both empty sets) regardless of a and b, as long as both a and b are >1/4. Therefore, even if a ≠ b, as long as both are >1/4, A and B are both empty and hence equal. Therefore, the equality A = B can occur in two scenarios:1. When a = b ≤ 1/4, leading to non-empty equal intervals.2. When a and b are both >1/4, leading to both being empty, regardless of their specific values.Therefore, if A = B, it is possible that a = b, or a ≠ b as long as both are >1/4. Hence, a = b is not a necessary condition for A = B. Therefore, the statement "Option A: a = b" is a sufficient condition for Option B: A = B, because if a = b, then regardless of the value of a, A and B will be the same (either both empty or both the same interval). However, it is not necessary, because A and B can also be equal when both are empty, even if a ≠ b. Therefore, the correct answer is A: Option A is a sufficient condition but not a necessary condition for option B. Wait, but let me check again. If a = b, then A = B. So, a = b is sufficient. But if A = B, it could be because a = b ≤ 1/4, or a and b are different but both >1/4. So, A = B can happen without a = b. Therefore, a = b is not necessary. Therefore, option A is a sufficient condition but not a necessary condition for option B. Therefore, answer is A. But let's look back at the options:A: Option A is a sufficient condition but not a necessary condition for option B.B: Option A is a necessary condition but not a sufficient condition for option B.C: Both.D: Neither.So, based on the reasoning above, since a = b is sufficient but not necessary, the answer is A. But wait, wait. Let me think again. If a and b are such that both are greater than 1/4, then A and B are both empty, hence equal. So, even if a ≠ b, as long as both are >1/4, A = B. Therefore, A = B does not require a = b. Therefore, a = b is not necessary for A = B. However, if a = b, then regardless of the value, A = B. Therefore, a = b is sufficient. So, sufficient but not necessary. Therefore, answer is A.But wait, in the first analysis, when A and B are non-empty, then A = B requires a = b. But when they are both empty, A = B even if a ≠ b. Therefore, in some cases, a ≠ b can still lead to A = B. Therefore, a = b is sufficient (because if you set a = b, you are guaranteed A = B) but not necessary (because you can have A = B without a = b, by having both a and b >1/4). Therefore, answer is A.Therefore, the correct answer should be A.Wait, but hold on. Let me check with specific examples.Example 1: Let a = 0 and b = 0. Then, A and B are both the solutions to x² - x ≤ 0. The roots are 0 and 1. So, the interval [0,1]. So, A = B.Example 2: Let a = 1/4 and b = 1/4. Then, quadratic becomes x² - x + 1/4 = 0, which factors to (x - 1/2)^2 = 0. So, the solution set is {1/2}, so A = B = {1/2}.Example 3: Let a = 1/2 and b = 1/2. Then, discriminant is 1 - 4*(1/2) = -1 < 0. So, A and B are both empty. Hence, A = B.Example 4: Let a = 1/2 and b = 3/4. Then, discriminant for A is 1 - 2 = -1, and for B is 1 - 3 = -2. Both A and B are empty. Therefore, A = B even though a ≠ b.Thus, in example 4, even though a ≠ b, A = B. Therefore, a = b is not necessary. However, if a = b, then A = B. So, sufficient but not necessary. So, answer is A.Therefore, the correct answer is option A.But let me check again. The original problem says "the correct answer is" with options A to D. The options are labeled A, B, C, D, but the options themselves are about whether option A (a = b) is a sufficient/necessary condition for option B (A = B). So, yes. The answer is A: Option A is a sufficient condition but not a necessary condition for option B.Final Answerboxed{A}</think>Given the sets ( A = { x in mathbb{R} mid x^2 - x + a leq 0 } ) and ( B = { x in mathbb{R} mid x^2 - x + b leq 0 } ), we need to determine the relationship between the conditions ( a = b ) (Option A) and ( A = B ) (Option B).1. Discriminant Analysis: - The discriminant of the quadratic equation ( x^2 - x + a = 0 ) is ( 1 - 4a ). - The discriminant of the quadratic equation ( x^2 - x + b = 0 ) is ( 1 - 4b ).2. Non-Empty Sets: - For ( A ) and ( B ) to be non-empty, their discriminants must be non-negative: ( 1 - 4a geq 0 ) and ( 1 - 4b geq 0 ). This implies ( a leq frac{1}{4} ) and ( b leq frac{1}{4} ).3. Roots and Intervals: - If ( a leq frac{1}{4} ), the roots of ( x^2 - x + a = 0 ) are ( frac{1 pm sqrt{1 - 4a}}{2} ), forming the interval ( left[ frac{1 - sqrt{1 - 4a}}{2}, frac{1 + sqrt{1 - 4a}}{2} right] ). - Similarly for ( b leq frac{1}{4} ), the interval for ( B ) is ( left[ frac{1 - sqrt{1 - 4b}}{2}, frac{1 + sqrt{1 - 4b}}{2} right] ).4. Equality of Sets: - If ( a = b ), the intervals for ( A ) and ( B ) are identical, making ( A = B ). This shows ( a = b ) is a sufficient condition. - However, if ( a neq b ) but both ( a ) and ( b ) are greater than ( frac{1}{4} ), both sets ( A ) and ( B ) are empty, hence equal. This shows ( a = b ) is not a necessary condition.5. Conclusion: - ( a = b ) is sufficient for ( A = B ) because equal ( a ) and ( b ) guarantee equal intervals or both sets being empty. - ( a = b ) is not necessary for ( A = B ) because different ( a ) and ( b ) can result in both sets being empty.Thus, the correct answer is (boxed{A}).]

🔑: Step-by-step Solution:# Sufficiency:To check if option A (a=b) is a sufficient condition for option B (A=B), let's assume a=b. Given the definitions of sets A and B:- A={xin mathbb{R} | x^{2}-x+aleqslant 0}- B={xin mathbb{R} | x^{2}-x+bleqslant 0}If a=b, we have:- For set A: x^{2}-x+aleqslant 0- For set B: x^{2}-x+bleqslant 0 Rightarrow x^{2}-x+aleqslant 0 (since a=b)Thus, the inequalities defining A and B are identical, which means A=B. This confirms that if a=b, then A=B, demonstrating that option A is a sufficient condition for option B.# Necessity:To check if option A (a=b) is a necessary condition for option B (A=B), we consider the case where A=B. Suppose both A and B are empty sets. This situation occurs when the discriminant of the quadratic equation x^2 - x + a = 0 and x^2 - x + b = 0 is less than 0, i.e., 1 - 4a < 0 and 1 - 4b < 0. For example, if we take a=1 and b=2, then:- For a=1, 1-4a<0 holds, making A an empty set.- For b=2, 1-4b<0 also holds, making B an empty set.Thus, A=B=varnothing, even though a neq b. This shows that a=b is not a necessary condition for A=B because we can have A=B without a being equal to b.# Conclusion:Based on the sufficiency and necessity analysis, option A (a=b) is a sufficient condition but not a necessary condition for option B (A=B).Therefore, the correct answer is boxed{text{A}}.

🧠:[<think>Okay, let's see. I need to figure out whether 1 is greater than the sum of these three fractions: 23/93, 41/165, and 71/143. Hmm, adding fractions can be tricky because they have different denominators. Let me think about how to approach this.First, maybe I can estimate each fraction to see if their sum is more or less than 1 without calculating the exact value. That might save some time. Let me look at each fraction individually.Starting with 23/93. Well, 23 is a prime number, and 93 divided by 3 is 31, so 93 is 3*31. Hmm, not sure if that helps. Let me divide 23 by 93. If I approximate 23/93, 93 goes into 23 zero times. Then 230 divided by 93 is about 2.47, so 0.247... So roughly 0.247. Let me check that with a calculator in my mind. 93 times 0.2 is 18.6, subtract that from 23, we get 4.4. 93 goes into 4.4 about 0.047 times. So yes, approximately 0.247. So about 0.25.Next, 41/165. Let's see. 165 is 5*33, which is 5*3*11. 41 divided by 165. Again, 165 goes into 41 zero times. 410 divided by 165 is 2.48, so 0.248. So similar to the first one. Let me check. 165*0.2=33, subtract that from 41, we have 8. 165 goes into 8 about 0.048 times. So approximately 0.248. So also about 0.25.Third fraction is 71/143. Hmm, 143 is 11*13. 71 divided by 143. Let's estimate. 71 is half of 142, so 71/143 is exactly 0.5. Wait, 143 divided by 2 is 71.5, so 71 is just slightly less than half. So 71/143 is approximately 0.496. So roughly 0.5.So adding up the estimates: 0.25 + 0.25 + 0.5 = 1.0. But wait, my approximations were rounding two of them down and one up? Let me check more accurately.First, 23/93. Let's compute this more precisely. 23 divided by 93. Let's do long division. 93 goes into 230 two times (2*93=186), remainder 44. Bring down a zero: 440. 93 goes into 440 four times (4*93=372), remainder 68. Bring down a zero: 680. 93 goes into 680 seven times (7*93=651), remainder 29. Bring down a zero: 290. 93 goes into 290 three times (3*93=279), remainder 11. Bring down a zero: 110. 93 goes into 110 once (1*93=93), remainder 17. So, so far, we have 0.24731... So approximately 0.2473.Second, 41/165. Let's divide 41 by 165. 165 goes into 410 two times (2*165=330), remainder 80. Bring down a zero: 800. 165 goes into 800 four times (4*165=660), remainder 140. Bring down a zero: 1400. 165 goes into 1400 eight times (8*165=1320), remainder 80. Wait, we had a remainder 80 before. So this will start repeating. So the decimal is 0.2484848... So approximately 0.2485.Third, 71/143. Let's divide 71 by 143. 143 goes into 710 four times (4*143=572), remainder 138. Bring down a zero: 1380. 143 goes into 1380 nine times (9*143=1287), remainder 93. Bring down a zero: 930. 143 goes into 930 six times (6*143=858), remainder 72. Bring down a zero: 720. 143 goes into 720 five times (5*143=715), remainder 5. Bring down a zero: 50. 143 goes into 50 zero times, remainder 50. Bring down another zero: 500. 143 goes into 500 three times (3*143=429), remainder 71. Bring down a zero: 710. Wait, we saw this before (710 was the first step after the decimal). So the decimal repeats. So the decimal is 0.4965034965... So approximately 0.4965.So adding the three decimal approximations:0.2473 (23/93)+0.2485 (41/165)+0.4965 (71/143)= Let's add them step by step.First, 0.2473 + 0.2485. Let's add the thousandths:0.2473+0.2485=0.4958Then add the third number, 0.4965:0.4958 + 0.4965. Let's line them up:0.4958+0.4965= ?Adding the ten-thousandths: 8 +5=13, so 3 carryover 1.Thousandths: 5+6=11 +1=12, so 2 carryover 1.Hundredths: 9+9=18 +1=19, so 9 carryover 1.Tenths: 4+4=8 +1=9.So total is 0.9923.Wait, so approximately 0.9923. So that's less than 1. But my approximations were precise to four decimal places. But perhaps the truncation or rounding errors might affect the result? Let me check.Wait, perhaps I should compute exact fractions. Maybe my decimal approximations are leading me slightly astray.Alternatively, maybe adding the fractions step by step.First, let's compute 23/93 + 41/165. To add these, find a common denominator. The denominators are 93 and 165.Factor them:93 = 3 * 31165 = 5 * 33 = 5 * 3 * 11So the least common multiple (LCM) is 3 * 5 * 11 * 31. Let's compute that:3*5=15; 15*11=165; 165*31=165*30 +165*1=4950 +165=5115.So LCM of 93 and 165 is 5115.So convert both fractions to denominator 5115.23/93 = (23 * (5115/93)) / 5115Compute 5115 ÷93. 93*50=4650, 5115-4650=465; 93*5=465. So 50+5=55. So 5115/93=55. Therefore, 23/93 = (23*55)/5115.23*55: 20*55=1100, 3*55=165, total=1100+165=1265. So 23/93 =1265/5115.Similarly, 41/165 = (41*(5115/165))/5115Compute 5115 ÷165. 165*30=4950, 5115-4950=165. So 30+1=31. Thus 5115/165=31.Therefore, 41/165=41*31 /5115. 41*31: 40*31=1240, 1*31=31, total=1271. So 41/165=1271/5115.Adding 1265 +1271=2536. So 2536/5115.Then adding 71/143 to that sum.Now, 2536/5115 +71/143. Find LCM of 5115 and 143.Factor 143: 11*13Factor 5115: previously, 5115=3*5*11*31So LCM is 3*5*11*31*13= 3*5=15, 15*11=165, 165*31=5115, 5115*13=66495Therefore, LCM=66495.Convert 2536/5115 to denominator 66495:2536/5115 = (2536*(66495/5115))/66495Compute 66495 ÷5115. Let's divide 66495 by 5115.5115*13=66495. So 66495/5115=13. Therefore, multiply numerator and denominator by 13.2536*13=2536*10 +2536*3=25360 +7608=32968So 2536/5115=32968/66495Similarly, 71/143 converted to denominator 66495:71/143 = (71*(66495/143))/66495Compute 66495 ÷143. 143*400=57200, 66495-57200=9295. 143*60=8580, 9295-8580=715. 143*5=715. So total 400+60+5=465. Thus, 66495/143=465.Therefore, 71/143=71*465 /66495Compute 71*465: 70*465=32550, 1*465=465, total=32550+465=33015So 71/143=33015/66495Now, adding 32968/66495 +33015/66495= (32968+33015)/66495=65983/66495So the total sum is 65983/66495.Now, let's compute this fraction. To check if it's greater than 1, but wait, the numerator is 65983 and denominator is 66495. Since 65983 <66495, this fraction is less than 1. Therefore, the sum is less than 1. Therefore, 1 is greater.But let me confirm my calculations because fractions can be error-prone.First, adding 23/93 and 41/165:We converted them to 1265/5115 and 1271/5115, which sum to 2536/5115. Then adding 71/143, converted to 33015/66495, and the previous sum converted to 32968/66495. Adding these gives 65983/66495. Let's check the numerator addition:32968 +33015. Let's add 32,968 + 33,015:32,968+33,015=65,983. Correct.And denominator is 66,495. So 65,983/66,495. Simplify this fraction.Let me see if there is a common factor. Let's check the difference between denominator and numerator: 66,495 -65,983=512. Hmm, 512 is 2^9. Let's check if 512 divides into both numerator and denominator.But 65,983 divided by 512? 512*128=65,536. 65,983-65,536=447. 447/512 is the remainder. So no.Check GCD of 65,983 and 66,495. Use Euclidean algorithm.GCD(66,495, 65,983) = GCD(65,983, 66,495 -65,983)= GCD(65,983, 512)Then GCD(65,983,512). Compute 65,983 ÷512.512*128=65,536. 65,983 -65,536=447.GCD(512,447). 512 ÷447=1 with remainder 65.GCD(447,65). 447 ÷65=6 with remainder 57.GCD(65,57). 65 ÷57=1 with remainder 8.GCD(57,8). 57 ÷8=7 with remainder 1.GCD(8,1). Which is 1. So GCD is 1. Therefore, the fraction reduces to 65,983/66,495, which is approximately 0.9923 as before.So yes, the sum is approximately 0.9923, which is less than 1. Therefore, 1 is greater.But let me cross-verify with another approach.Alternative method: Instead of adding all three fractions, compare the sum to 1 by subtracting each fraction from 1 step by step.Start with 1. Subtract 23/93: 1 -23/93= (93-23)/93=70/93≈0.7527.Then subtract 41/165 from that: 70/93 -41/165. Let's compute this.Convert to common denominator, which we know is 5115.70/93=70*55/5115=3850/511541/165=41*31/5115=1271/5115So 3850 -1271=2579. So 2579/5115≈0.5044.Then subtract the last fraction, 71/143: 2579/5115 -71/143.Convert to common denominator 66495.2579/5115=2579*13/66495=33,527/66,49571/143=71*465/66495=33,015/66,495Subtracting: 33,527 -33,015=512. So 512/66,495≈0.0077.Therefore, 1 - (sum of fractions)=512/66495≈0.0077, so the sum is 1 -0.0077≈0.9923. Same result as before. Hence, the sum is approximately 0.9923, which is less than 1. Therefore, 1 is greater.Alternatively, another way to check is cross-multiplying to compare the sum with 1.We can check if 23/93 +41/165 +71/143 >1.Multiply both sides by 93*165*143 to eliminate denominators. Since all denominators are positive, inequality direction remains same.So:23*165*143 +41*93*143 +71*93*165 >? 93*165*143Compute left side:First term:23*165*143Second term:41*93*143Third term:71*93*165Right side:93*165*143So subtract right side from left side:23*165*143 +41*93*143 +71*93*165 -93*165*143Factor terms:Let's see, the first term is 23*(165*143), the second is 41*93*143, third is 71*93*165, subtract 93*165*143.Factor out 93*165*143 from first and fourth term:(23 -93)*165*143 +41*93*143 +71*93*165Wait, maybe not. Alternatively, compute each term:Compute 23*165*143:23*165=3795; 3795*143. Let's compute 3795*100=379,500; 3795*40=151,800; 3795*3=11,385. Sum these: 379,500 +151,800=531,300 +11,385=542,685. So first term is 542,685.Second term:41*93*143.41*93=3,813; 3,813*143. Compute 3,813*100=381,300; 3,813*40=152,520; 3,813*3=11,439. Sum:381,300 +152,520=533,820 +11,439=545,259. Second term is 545,259.Third term:71*93*165.71*93=6,603; 6,603*165. Compute 6,603*100=660,300; 6,603*60=396,180; 6,603*5=33,015. Sum:660,300 +396,180=1,056,480 +33,015=1,089,495. Third term is 1,089,495.Right side:93*165*143.93*165=15,345; 15,345*143. Compute 15,345*100=1,534,500; 15,345*40=613,800; 15,345*3=46,035. Sum:1,534,500 +613,800=2,148,300 +46,035=2,194,335.Now, left side total:542,685 +545,259 +1,089,495.Compute step by step:542,685 +545,259 =1,087,9441,087,944 +1,089,495=2,177,439Right side is 2,194,335.So left side is 2,177,439. Compare to right side 2,194,335. So left side is less than right side by 2,194,335 -2,177,439=16,896.Therefore, 23/93 +41/165 +71/143 <1.Hence, 1 is greater.Therefore, after multiple methods: decimal approximation, exact fraction addition, and cross-multiplication, all confirm that the sum is less than 1. So the answer is 1 is greater.Final Answerboxed{1}</think>To determine whether 1 is greater than the sum of the fractions (frac{23}{93} + frac{41}{165} + frac{71}{143}), we can use several methods to verify the result.1. Decimal Approximation: - (frac{23}{93} approx 0.2473) - (frac{41}{165} approx 0.2485) - (frac{71}{143} approx 0.4965) - Summing these approximations: (0.2473 + 0.2485 + 0.4965 approx 0.9923), which is less than 1.2. Exact Fraction Addition: - Convert each fraction to a common denominator and add them step-by-step: - (frac{23}{93} + frac{41}{165} = frac{2536}{5115}) - Adding (frac{71}{143}) results in (frac{65983}{66495}) - The fraction (frac{65983}{66495}) is less than 1 because the numerator (65983) is less than the denominator (66495).3. Cross-Multiplication Method: - Compare the sum of the fractions to 1 by cross-multiplying: - Calculate each term and subtract the product of the denominators from the sum of the products of the numerators: - The result shows that the left side (sum of the fractions) is less than the right side (1).After verifying through multiple methods, we conclude that the sum (frac{23}{93} + frac{41}{165} + frac{71}{143}) is less than 1. Therefore, the answer is:(boxed{1}),<think>Okay, let's see. I need to figure out whether 1 is greater than the sum of these three fractions: 23/93, 41/165, and 71/143. Hmm, adding fractions can be tricky, especially with different denominators. Maybe I should start by estimating each fraction to see if their sum is more or less than 1. First, let's look at 23/93. Let me divide 23 by 93. Since 23 is less than a third of 93 (since 93 divided by 3 is 31), so 23/93 is a bit less than 1/3. Let me check that. 1/3 is approximately 0.333... So 23 divided by 93. Let's do the division: 93 goes into 23 zero times. Add a decimal point. 93 goes into 230 two times (since 2*93=186). Subtract 186 from 230, we get 44. Bring down the next 0: 440. 93 goes into 440 four times (4*93=372). Subtract 372 from 440, get 68. Bring down a 0: 680. 93 goes into 680 seven times (7*93=651). Subtract 651 from 680, get 29. Bring down a 0: 290. 93 goes into 290 three times (3*93=279). Subtract 279 from 290, get 11. So, so far, 23/93 is approximately 0.2473... Hmm, so roughly 0.247.Next, 41/165. Let's divide 41 by 165. 165 goes into 41 zero times. 41.000 divided by 165. 165*0.2=33.0. So 0.2 is 33.0. Subtract that from 41, get 8.0. Bring down a zero: 80. 165 goes into 80 zero times. Bring down another zero: 800. 165*4=660. So 4*165=660. Subtract that from 800, get 140. Bring down a zero: 1400. 165*8=1320. Subtract that, get 80. Wait, this is starting to repeat. So 41/165 is 0.24848..., approximately 0.248.Third fraction: 71/143. Let's divide 71 by 143. Again, 143 is a bit more than 140. 71 divided by 143. Let's see, 143*0.5=71.5. So 0.5 is 71.5, which is just a bit more than 71, so 71/143 is just a bit less than 0.5. Let's calculate. 143 goes into 71.0 zero times. 143 goes into 710 four times (4*143=572). Subtract 572 from 710, get 138. Bring down a zero: 1380. 143*9=1287. Subtract 1287 from 1380, get 93. Bring down a zero: 930. 143*6=858. Subtract 858 from 930, get 72. Bring down a zero: 720. 143*5=715. Subtract 715 from 720, get 5. So, it's approximately 0.4965... So roughly 0.496.Adding them up: 0.247 + 0.248 + 0.496. Let's see. 0.247 + 0.248 is 0.495. Then add 0.496 gives us 0.991. Hmm, so that's approximately 0.991, which is just under 1. So 1 is greater? But wait, maybe my approximations are off. Because when I approximated each decimal, I might have truncated some parts, so maybe the actual sum is slightly higher?Let me check each fraction again with more precision.Starting with 23/93. Earlier, we had it as approximately 0.24731... So 0.2473.41/165 was approximately 0.24848..., so 0.2485.71/143 was approximately 0.49650..., so 0.4965.Adding these: 0.2473 + 0.2485 = 0.4958. Then add 0.4965: 0.4958 + 0.4965 = 0.9923. So approximately 0.9923. That's still less than 1 by about 0.0077. So maybe 1 is still greater.But maybe if I calculate the exact sum. Let's find a common denominator for these fractions. The denominators are 93, 165, and 143. Let's factor each one:93: 3 * 31165: 5 * 33 = 5 * 3 * 11143: 11 * 13So the prime factors involved are 3, 5, 11, 13, and 31. The least common denominator (LCD) would be the product of the highest powers of all primes: 3^1, 5^1, 11^1, 13^1, 31^1. So LCD = 3 * 5 * 11 * 13 * 31. Let me calculate that step by step:First, 3 * 5 = 1515 * 11 = 165165 * 13 = 21452145 * 31. Let's compute 2145*30=64,350 and add 2145: 64,350 + 2,145 = 66,495So the LCD is 66,495.Now, convert each fraction to have this denominator:First fraction: 23/93. Multiply numerator and denominator by (66,495 / 93) = (66,495 ÷ 93). Let's compute 93 * 700 = 65,100. 66,495 - 65,100 = 1,395. 93 * 15 = 1,395. So 700 + 15 = 715. So 66,495 ÷ 93 = 715. Therefore, 23/93 = 23*715 / 66,495. Let's compute 23*700=16,100 and 23*15=345. So total 16,100 + 345 = 16,445. So 16,445/66,495.Second fraction: 41/165. Multiply numerator and denominator by (66,495 / 165). 165 * 400 = 66,000. 66,495 - 66,000 = 495. 165 * 3 = 495. So 400 + 3 = 403. So 66,495 ÷ 165 = 403. Therefore, 41/165 = 41*403 / 66,495. Let's compute 40*403 = 16,120 and 1*403=403. Total 16,120 + 403 = 16,523. So 16,523/66,495.Third fraction: 71/143. Multiply numerator and denominator by (66,495 / 143). Let's compute 143 * 400 = 57,200. 66,495 - 57,200 = 9,295. 143 * 65 = 9,295 (since 143*60=8,580 and 143*5=715; 8,580+715=9,295). So 400 + 65 = 465. So 66,495 ÷ 143 = 465. Therefore, 71/143 = 71*465 / 66,495. Let's compute 70*465=32,550 and 1*465=465. So total 32,550 + 465 = 33,015. So 33,015/66,495.Now, sum the numerators: 16,445 + 16,523 + 33,015. Let's add them step by step.16,445 + 16,523 = 32,968. Then 32,968 + 33,015. 32,968 + 33,000 = 65,968. Then add 15: 65,983. So total numerator is 65,983. So the sum is 65,983 / 66,495.So, 65,983 divided by 66,495. Let's compute this. Subtract numerator from denominator: 66,495 - 65,983 = 512. So 65,983 / 66,495 = 1 - 512/66,495. So, the fraction is 512/66,495 less than 1. To see how much that is, let's compute 512/66,495. Let's divide numerator and denominator by 16: 512 ÷ 16 = 32; 66,495 ÷ 16 = 4,155.9375. Hmm, not helpful. Maybe approximate it as decimal.Divide 512 by 66,495. Let's see:66,495 goes into 512 zero times. Add decimal: 512.0 is 5120.0 divided by 66,495. 66,495 * 0.07 is 4,654.65. Subtract that from 5,120: 5,120 - 4,654.65 = 465.35. Bring down a zero: 4,653.5. 66,495 * 0.0007 is 46.5465. Subtract that from 4,653.5: 4,653.5 - 46.5465 = 4,606.9535. This is getting complicated. Alternatively, maybe approximate as 512/66,495 ≈ 512/66,500 ≈ 0.0077 (since 66,500 * 0.007 = 465.5, 66,500*0.0077=512.05). So approximately 0.0077. So the sum is 1 - 0.0077 ≈ 0.9923, which matches our earlier approximation.Therefore, the sum of the three fractions is approximately 0.9923, which is less than 1. Therefore, 1 is greater.Wait a second, but the problem asks "What is greater: 1 or the sum?" So 1 is greater. But let me double-check because sometimes these fraction problems can have a closer call. Let me confirm the exact value.We have 65,983/66,495. Let's compute this as a decimal more precisely. Let's do long division for 65,983 ÷ 66,495.65,983 divided by 66,495. Since 65,983 is less than 66,495, the result is less than 1. Let's compute it as 0. (something). Let's set it up as 65,983.0000 divided by 66,495.Multiply numerator and denominator by 10 to make it 659,830 ÷ 664,95.Wait, maybe it's easier to note that 65,983 / 66,495 = (66,495 - 512)/66,495 = 1 - 512/66,495.So 512 divided by 66,495. Let's compute 512 ÷ 66,495.66,495 goes into 512 0 times. 66,495 goes into 5120 0 times. 66,495 goes into 51200 0 times. 66,495 goes into 512000 approximately 7 times (7*66,495=465,465). Subtract 465,465 from 512,000: 512,000 - 465,465 = 46,535. Bring down the next 0: 465,350. 66,495 goes into that 7 times again (7*66,495=465,465). Wait, but 465,350 is less than 465,465. So 6 times: 6*66,495=398,970. Subtract from 465,350: 465,350 - 398,970 = 66,380. Bring down a 0: 663,800. 66,495 goes into that 9 times (9*66,495=598,455). Subtract: 663,800 - 598,455 = 65,345. Bring down a 0: 653,450. 66,495 goes into that 9 times again: 598,455. Subtract: 653,450 - 598,455 = 54,995. Bring down a 0: 549,950. 66,495 goes into that 8 times (8*66,495=531,960). Subtract: 549,950 - 531,960 = 17,990. Bring down a 0: 179,900. 66,495 goes into that 2 times (2*66,495=132,990). Subtract: 179,900 - 132,990 = 46,910. Bring down a 0: 469,100. 66,495 goes into that 7 times (7*66,495=465,465). Subtract: 469,100 - 465,465 = 3,635. At this point, we can see that it's starting to repeat or continue, but the decimal is approximately 0.0077.So, 512/66,495 ≈ 0.0077, so the total sum is 1 - 0.0077 ≈ 0.9923. Therefore, 0.9923 is indeed less than 1, so 1 is greater.But wait, just to be thorough, is there a chance that my exact fractions when summed might actually be slightly higher? Let me check the numerators again:First fraction: 23*715 = 16,445. Let's confirm 23*700=16,100 and 23*15=345, so 16,100+345=16,445. Correct.Second fraction: 41*403=16,523. 40*403=16,120 and 1*403=403, so 16,120+403=16,523. Correct.Third fraction: 71*465=33,015. 70*465=32,550 and 1*465=465, so 32,550+465=33,015. Correct.Adding all numerators: 16,445 + 16,523 = 32,968. 32,968 + 33,015 = 65,983. Correct.Denominator: 66,495. Correct.So 65,983/66,495 is indeed the exact sum. So converting that to decimal, as we saw, is approximately 0.9923. So yes, less than 1.Therefore, the answer is that 1 is greater.Final Answerboxed{1}</think>To determine whether 1 is greater than the sum of the fractions (frac{23}{93} + frac{41}{165} + frac{71}{143}), we start by approximating each fraction:1. (frac{23}{93}) is approximately 0.2473.2. (frac{41}{165}) is approximately 0.2485.3. (frac{71}{143}) is approximately 0.4965.Adding these approximations:[0.2473 + 0.2485 + 0.4965 approx 0.9923]This suggests the sum is approximately 0.9923, which is less than 1. To confirm, we calculate the exact sum by finding a common denominator. The denominators are 93, 165, and 143, which factor into primes as follows:- 93 = 3 × 31- 165 = 3 × 5 × 11- 143 = 11 × 13The least common denominator (LCD) is (3 times 5 times 11 times 13 times 31 = 66,495).Converting each fraction:- (frac{23}{93} = frac{16,445}{66,495})- (frac{41}{165} = frac{16,523}{66,495})- (frac{71}{143} = frac{33,015}{66,495})Summing the numerators:[16,445 + 16,523 + 33,015 = 65,983]Thus, the exact sum is (frac{65,983}{66,495}). Converting this to a decimal:[frac{65,983}{66,495} approx 0.9923]Since 0.9923 is less than 1, we conclude that 1 is greater.[boxed{1}]]

🔑:Answer: One is greater.## First solution.frac{23}{93}+frac{41}{165}+frac{71}{143}<frac{23}{92}+frac{41}{164}+frac{71}{142}=frac{1}{4}+frac{1}{4}+frac{1}{2}=1 .## Second solution.begin{aligned}frac{23}{93}+frac{41}{165}+frac{71}{143} & =frac{23 cdot 165 cdot 143+93 cdot 41 cdot 143+93 cdot 165 cdot 71}{93 cdot 165 cdot 143}= & =frac{542685+545259+1089495}{2194335}=frac{2177439}{2194335}left[=frac{65983}{66495}right]<1end{aligned}

🧠:[<think>Okay, so I need to figure out how much someone's monthly salary is, given that they paid 30 yuan in taxes in January according to China's Personal Income Tax Law. Let me start by understanding the tax brackets here.First, the problem says that citizens don't have to pay tax on the first 800 yuan of their monthly salary. Anything above 800 is taxable. The tax is calculated in segments, kind of like how progressive tax systems work in other countries. The table provided shows the different tax rates for different portions of the taxable income. The brackets are:1. Up to 500 yuan taxed at 5%.2. The portion from 500 to 2000 yuan taxed at 10%.3. The portion from 2000 to 5000 yuan taxed at 15%.4. And so on...But the person's tax is only 30 yuan. So, their taxable income must be in one of the lower brackets. Let me break this down step by step.First, let's define the person's total salary as S. The taxable income would then be S - 800, since the first 800 yuan is tax-free. The tax is calculated on this taxable income in segments.Given that the tax paid is 30 yuan, which is relatively low, it's likely that their taxable income falls within the first two brackets. Let's check.The first bracket is up to 500 yuan taxed at 5%. So, if someone's taxable income is 500 yuan, the tax would be 500 * 5% = 25 yuan. Then, the next bracket is from 500 to 2000 yuan taxed at 10%. But the total tax here is 30 yuan. Since 25 yuan is the maximum tax in the first bracket, the remaining 5 yuan must come from the next bracket.So, let's compute how much income would result in 5 yuan tax at the 10% rate. Since 10% of some amount equals 5 yuan, that amount is 5 / 0.10 = 50 yuan. Therefore, the taxable income would be 500 (from the first bracket) + 50 (from the second bracket) = 550 yuan.Adding this to the non-taxable 800 yuan, the total salary would be 800 + 550 = 1350 yuan.Wait, let me verify that again. If the taxable income is 550 yuan, then the first 500 is taxed at 5%, so 500 * 0.05 = 25 yuan. The remaining 50 yuan is taxed at 10%, so 50 * 0.10 = 5 yuan. Total tax is 25 + 5 = 30 yuan. That checks out. So the total salary would be 800 + 550 = 1350 yuan.But just to make sure I didn't skip any steps or make any miscalculations. Let's think again. Suppose the person earned 1350 yuan. Subtract 800, taxable income is 550. First 500 taxed at 5%: 25. The next 50 taxed at 10%: 5. Total tax 30. Perfect. So 1350 yuan is the answer?Alternatively, maybe there's another way to approach this. Let me consider the tax brackets again. The first 500 taxed at 5% gives a maximum of 25. If someone's tax is 30, which is 5 more than 25, that must be from the next bracket. So, 5 / 0.10 = 50. So, total taxable income is 500 + 50 = 550. Total salary is 800 + 550 = 1350. Yeah, that seems consistent.I don't think there's a possibility here that the tax could be in a higher bracket. For example, if someone was in the 15% bracket, their tax would be even higher. Let's test that. Suppose taxable income is 2000. Then tax would be 500*5% + (2000 - 500)*10% = 25 + 150 = 175. That's way higher than 30. So definitely, the person is in the first two brackets.Alternatively, maybe they are entirely in the first bracket. If taxable income is X, then 5% of X = 30. Then X = 30 / 0.05 = 600. But 600 is more than 500, which is the limit for the first bracket. So that's not possible. So the tax has to be split between the first and second brackets. Which brings us back to 25 + 5 = 30. So yes, 500 and 50.Therefore, total taxable income is 550, total salary is 800 + 550 = 1350 yuan.I think that's the correct answer. Let me just check if there's another possible interpretation. For example, is the "monthly taxable income" the amount after subtracting 800, and then each bracket is applied to the amount over the previous bracket? Yes, that's standard for progressive taxes.So, for example, if someone earns 1350:Taxable income: 1350 - 800 = 550.First 500 taxed at 5%: 25.Next 50 (since 550 - 500 = 50) taxed at 10%: 5.Total tax: 30. Perfect.Alternatively, if someone earned, say, 1300. Taxable income 500. Tax 25. So 1300 would have tax 25. But the person in the problem has tax 30. So 1300 is too low. 1350 is the correct.Alternatively, 1400. Taxable income 600. Tax would be 25 + (100 * 10%) = 25 + 10 = 35. So tax would be 35. So 1400 would have 35 tax. So 1350 gives 30. Therefore, the answer is 1350.Yes, this makes sense. I don't see any mistakes in the reasoning here.Final Answerboxed{1350}</think>Given that a person's tax for January is 30 yuan, we need to determine his monthly salary and wages. According to China's Personal Income Tax Law, the first 800 yuan is tax-free. The taxable income is the amount exceeding 800 yuan, which is taxed progressively.1. The first 500 yuan of taxable income is taxed at 5%. The maximum tax for this portion is: [ 500 times 0.05 = 25 text{ yuan} ]2. The remaining tax of 30 yuan - 25 yuan = 5 yuan must come from the next bracket, which is taxed at 10%. The amount taxed at 10% is: [ frac{5}{0.10} = 50 text{ yuan} ]3. Therefore, the total taxable income is: [ 500 + 50 = 550 text{ yuan} ]4. Adding the non-taxable portion of 800 yuan, the total monthly salary is: [ 800 + 550 = 1350 text{ yuan} ]Verification:- Taxable income: 1350 - 800 = 550 yuan- First 500 yuan taxed at 5%: (500 times 0.05 = 25) yuan- Next 50 yuan taxed at 10%: (50 times 0.10 = 5) yuan- Total tax: 25 + 5 = 30 yuanThus, the monthly salary and wages are (boxed{1350}) yuan.,<think>Okay, so I need to figure out someone's monthly salary based on the tax they paid, which is 30 yuan. Let me start by recalling how the Chinese personal income tax works according to the problem. The first 800 yuan is tax-free. Anything above 800 is taxable, and that taxable amount is divided into segments taxed at different rates. The table provided shows that the first 500 yuan over 800 is taxed at 5%, the next 1500 yuan (from 500 to 2000) is taxed at 10%, and then up to 5000 it's 15%, and so on. Since the tax paid is 30 yuan, I need to figure out which tax bracket this falls into. Let me break it down step by step. First, the taxable income is the salary minus 800 yuan. Let's denote the total salary as S. Therefore, taxable income is (S - 800). The tax is calculated on this amount in segments. The first segment is up to 500 yuan, taxed at 5%. If the taxable income is within this first bracket, the tax would be 5% of (S - 800). If it's more than 500, then the first 500 is taxed at 5%, and the amount over 500 up to 2000 is taxed at 10%, and so on.Given that the tax is 30 yuan, let's first check if the taxable income is within the first bracket. If all the tax came from the first bracket, then 5% of taxable income = 30 yuan. So, 0.05 * taxable income = 30. Therefore, taxable income would be 30 / 0.05 = 600 yuan. Wait, but the first bracket is up to 500 yuan. So if taxable income is 600, that's 500 in the first bracket and 100 in the second bracket. So the tax would be 500 * 0.05 + 100 * 0.10 = 25 + 10 = 35 yuan. But the tax paid is 30 yuan, which is less than 35. Therefore, the taxable income must be somewhere in the first bracket but let's verify.Wait, maybe I made a mistake here. If the taxable income is 600, which is over 500, so the tax calculation would be 500*5% + (600-500)*10% = 25 + 10 = 35, as above. So if the tax is 30, then maybe the taxable income is such that the entire tax is from the first bracket. But 5% of taxable income = 30. So taxable income = 30 / 0.05 = 600. But that 600 is over 500, so that's actually in the second bracket. That's a contradiction. Therefore, this suggests that the taxable income is such that part is taxed at 5% and part at 10%, but then the total tax would be more than 25 (from the first 500). But 30 is only 5 more than 25. So perhaps the taxable income is 500 (taxed at 5%) giving 25 yuan tax, and then the remaining 5 yuan tax comes from the next bracket. Let's check.If tax is 30, then 25 yuan comes from the first 500, and the remaining 5 yuan is from the next bracket taxed at 10%. So 5 / 0.10 = 50 yuan. Therefore, the taxable income is 500 + 50 = 550 yuan. Therefore, total salary would be 800 + 550 = 1350 yuan. Let's verify this:Taxable income: 1350 - 800 = 550.First 500 taxed at 5%: 500 * 0.05 = 25.Next 50 taxed at 10%: 50 * 0.10 = 5.Total tax: 25 + 5 = 30. Perfect. So the total salary is 1350 yuan.But wait, let me make sure there isn't another way. Suppose the entire taxable income was in the first bracket. Then taxable income would be 30 / 0.05 = 600. But 600 is more than 500, so the tax should be calculated as per the brackets, which would require part of it to be taxed at 10%, leading to higher tax. Hence, the taxable income can't be 600. Therefore, the only possible way is that the taxable income is 550, leading to 25 + 5 = 30 tax. Therefore, the total salary is 800 + 550 = 1350 yuan. Therefore, the answer is 1350.Wait, but let me confirm once again. Let's compute step by step:Total salary = 1350.Taxable income = 1350 - 800 = 550.First 500 yuan taxed at 5%: 500 * 0.05 = 25.Remaining 550 - 500 = 50 yuan taxed at 10%: 50 * 0.10 = 5.Total tax: 25 + 5 = 30. Exactly matches the given tax. So yes, 1350 is correct.But hold on, maybe the person's salary is such that even after subtracting 800, the taxable income is exactly 500? Then tax would be 500*0.05=25 yuan. But the tax paid is 30, which is more than 25. Therefore, it's definitely in the next bracket. So the taxable income is between 500 and 2000. Therefore, the tax is 25 + (taxable income - 500)*0.10. Let's set that equal to 30.25 + (taxable income - 500)*0.10 = 30.Subtract 25: (taxable income - 500)*0.10 = 5.Divide by 0.10: taxable income - 500 = 50.Therefore, taxable income = 550. Therefore, total salary is 800 + 550 = 1350.Yes, that seems consistent. Therefore, the answer should be 1350 yuan.Final AnswerThe person's monthly salary and wages for that month are boxed{1350} yuan.</think>To determine the monthly salary based on the tax paid, we start by noting that the tax-free portion is 800 yuan. The tax paid is 30 yuan, which must be calculated from the taxable income exceeding 800 yuan. The tax rates are 5% for the first 500 yuan and 10% for the portion exceeding 500 yuan up to 2000 yuan.Given the tax amount of 30 yuan, we check if it falls into the second tax bracket. The first 500 yuan of taxable income is taxed at 5%, resulting in 25 yuan. The remaining tax (30 yuan - 25 yuan = 5 yuan) must come from the next bracket taxed at 10%. Calculating the portion taxed at 10%:[5 , text{yuan} = 0.10 times text{amount} implies text{amount} = frac{5}{0.10} = 50 , text{yuan}]Thus, the total taxable income is:[500 , text{yuan} + 50 , text{yuan} = 550 , text{yuan}]Adding the tax-free portion:[800 , text{yuan} + 550 , text{yuan} = 1350 , text{yuan}]Verification:- Taxable income: (1350 - 800 = 550 , text{yuan})- Tax on first 500 yuan: (500 times 0.05 = 25 , text{yuan})- Tax on remaining 50 yuan: (50 times 0.10 = 5 , text{yuan})- Total tax: (25 + 5 = 30 , text{yuan})Therefore, the person's monthly salary and wages for that month are (boxed{1350}) yuan.]

🔑:=、 11.1350

🧠:[<think>Okay, let's tackle this problem step by step. So, we have the function f(x) = (1 - x)/(a x) + ln x. There are two parts: part I asks for the maximum and minimum values when a = 1/2, on the interval [1, e]. Part II is about finding the range of positive a such that g(x) = f(x) - (1/4)x is increasing on [1, e]. Let's start with part I.First, when a = 1/2, substitute that into the function. So f(x) becomes (1 - x)/( (1/2)x ) + ln x. Let me compute that. The denominator is (1/2)x, so dividing by that is the same as multiplying by 2/x. Therefore, (1 - x) * 2 / x + ln x. Simplify that: 2(1 - x)/x + ln x. Which is 2*(1/x - 1) + ln x. So 2/x - 2 + ln x. So f(x) simplifies to 2/x + ln x - 2.Now, to find the maximum and minimum on [1, e], we need to find critical points by taking the derivative and evaluating endpoints.Compute f'(x). The derivative of 2/x is -2/x². The derivative of ln x is 1/x. The derivative of -2 is 0. So f'(x) = -2/x² + 1/x. Let me write that as f'(x) = ( -2 + x ) / x². Because combining the terms: -2/x² + 1/x = (-2 + x)/x². So f'(x) = (x - 2)/x².Set derivative equal to zero: (x - 2)/x² = 0. The numerator must be zero, so x - 2 = 0 => x = 2. So critical point at x = 2. Now, check if x = 2 is in the interval [1, e]. Since e is approximately 2.718, so 2 is within [1, e]. Therefore, critical points are x = 2, and endpoints x = 1 and x = e.Now evaluate f(x) at x = 1, x = 2, and x = e.First, f(1): 2/1 + ln 1 - 2 = 2 + 0 - 2 = 0.Next, f(2): 2/2 + ln 2 - 2 = 1 + ln 2 - 2 = (ln 2) -1 ≈ 0.6931 -1 ≈ -0.3069.Then f(e): 2/e + ln e - 2 = 2/e + 1 - 2 ≈ 0.7358 +1 -2 ≈ -0.2642.So f(1) = 0, f(2) ≈ -0.3069, f(e) ≈ -0.2642.Comparing these values, the maximum is at x=1 with value 0, and the minimum is at x=2 with approximately -0.3069. But we need to write exact values, not approximations. For f(2), exact value is (ln 2) -1. For f(e), exact value is 2/e +1 -2 = 2/e -1. Let me check that again. Wait, f(e) is 2/e + ln e -2. Since ln e =1, so 2/e +1 -2 = 2/e -1. Similarly, f(2) is 1 + ln2 -2 = ln2 -1. So exact expressions are:Maximum value: 0 at x=1.Minimum value: ln2 -1 at x=2.But let me confirm if there are any other critical points. The derivative is (x -2)/x². So only x=2 is critical point. Therefore, yes, only three points to check. So part I answer: maximum 0 at x=1, minimum ln2 -1 at x=2.Now part II: Find the range of positive a such that g(x) = f(x) - (1/4)x is increasing on [1,e]. So g(x) is increasing if its derivative is non-negative on [1,e].First, compute g(x). Since f(x) is (1 -x)/(a x) + ln x, then g(x) = (1 -x)/(a x) + ln x - (1/4)x.Compute the derivative g’(x). Let's compute derivative term by term.First term: (1 - x)/(a x). Let me rewrite this as (1/(a x)) - (x)/(a x) = 1/(a x) - 1/a. Therefore, derivative of first term is -1/(a x²) - 0, because derivative of 1/(a x) is -1/(a x²), and derivative of -1/a is 0. So first term derivative: -1/(a x²).Second term: ln x, derivative is 1/x.Third term: -(1/4)x, derivative is -1/4.So overall, g’(x) = -1/(a x²) + 1/x -1/4.We need g’(x) ≥ 0 for all x in [1,e].Therefore, the inequality is: -1/(a x²) + 1/x -1/4 ≥ 0 for all x in [1,e].We can rearrange terms:-1/(a x²) + 1/x -1/4 ≥ 0Multiply both sides by a x² (since a is positive and x² is positive, inequality sign remains):-1 + a x - (a x²)/4 ≥ 0Which simplifies to:a x - (a x²)/4 -1 ≥ 0Factor out a:a [x - x²/4] -1 ≥ 0Alternatively, write:a(x - x²/4) ≥1So, a ≥ 1 / (x - x²/4 )Therefore, to satisfy this inequality for all x in [1,e], the value of a must be greater than or equal to the maximum value of 1/(x - x²/4) on [1,e].Therefore, the minimal a required is the maximum of 1/(x - x²/4) over x in [1,e], so a ≥ max_{x ∈ [1,e]} [1/(x - x²/4)].So our task reduces to finding the maximum of the function h(x) = 1/(x - (x²)/4) on [1,e], then a ≥ that maximum.But let's analyze h(x) = 1/(x - x²/4). Let me first simplify the denominator:x - x²/4 = (4x - x²)/4 = x(4 - x)/4.Thus, h(x) = 1 / [x(4 - x)/4] = 4 / [x(4 - x)].Therefore, h(x) = 4 / [x(4 - x)].So now, we need to find the maximum of h(x) = 4 / [x(4 - x)] on [1, e].To find the maximum of h(x), since h(x) is continuous on [1, e], we can find its critical points and evaluate at endpoints.First, compute derivative h’(x):Let h(x) = 4 / [x(4 - x)] = 4 / (4x - x²). Let me write it as h(x) = 4*(4x - x²)^(-1).Derivative h’(x) = 4*(-1)*(4x - x²)^(-2)*(4 - 2x) = -4*(4 - 2x)/(4x - x²)^2.Set derivative equal to zero:-4*(4 - 2x)/(4x - x²)^2 = 0.The numerator must be zero: 4 - 2x = 0 => x = 2.So critical point at x = 2. Check if x=2 is in [1, e]. Since e ≈2.718, yes, x=2 is in [1, e].Therefore, critical points are x=2 and endpoints x=1 and x=e.Compute h(x) at these points:h(1) = 4 / [1*(4 -1)] = 4/(1*3) = 4/3 ≈1.333.h(2) = 4 / [2*(4 -2)] = 4/(2*2) = 4/4 =1.h(e) =4/[e*(4 - e)] ≈4/[2.718*(4 -2.718)] ≈4/[2.718*1.282] ≈4/3.483 ≈1.148.Therefore, h(x) reaches maximum at x=1 with h(1)=4/3, then decreases to h(2)=1, then increases slightly to h(e)≈1.148. Wait, but wait, h(e) is approximately 1.148 which is less than 4/3≈1.333. So maximum is at x=1.Wait, but let me compute h(e) exactly:h(e) =4/[e*(4 - e)].But 4 - e is approximately 1.282, and e is approximately 2.718, so e*(4 -e) ≈2.718*1.282≈3.483, so 4/3.483≈1.148, which is less than 4/3≈1.333. Therefore, the maximum of h(x) on [1,e] is at x=1, which is 4/3.Hence, the maximum value of h(x) on [1,e] is 4/3. Therefore, a must be ≥4/3.But wait, let me check the calculation again. The original expression was:a ≥ 1/(x -x²/4). Then h(x)=1/(x -x²/4). But then we simplified h(x) as 4/[x(4 -x)]. Let me check that step again.Original: x -x²/4. Let me factor x:x(1 -x/4). Alternatively, x - x²/4 = (4x -x²)/4 = x(4 -x)/4. Therefore, 1/(x -x²/4) = 4/(x(4 -x)), which is h(x). So yes, that's correct.Therefore, h(x)=4/(x(4 -x)). So maximum at x=1:4/(1*3)=4/3, which is correct. Then h(2)=4/(2*2)=1, and h(e)=4/(e*(4 -e)).Since 4 - e ≈1.282, e*(4 -e)≈2.718*1.282≈3.483. So 4 divided by that is approximately 1.148. So yes, 4/3≈1.333 is the maximum. So the maximum of h(x) is 4/3 at x=1.Therefore, a must be ≥4/3. So the range of a is [4/3, ∞).But let's double-check the inequality:We had to ensure that g’(x) ≥0 for all x in [1,e]. Which transformed into a ≥4/(x(4 -x)). The maximum of the right-hand side is 4/3. Therefore, a must be at least 4/3. So positive real numbers a with a ≥4/3.Hence, the answer for part II is a ≥4/3.Wait, but let's confirm with another approach. Suppose a =4/3. Then check if g’(x) ≥0.Compute g’(x) when a=4/3:g’(x)= -1/(a x²) +1/x -1/4.Substitute a=4/3:-1/( (4/3)x² ) +1/x -1/4 = -3/(4x²) +1/x -1/4.Combine terms:Let's write all terms with denominator 4x²:-3/(4x²) + 4x/(4x²) - (x²)/(4x²) =Wait, perhaps better to check at critical points. Let me check at x=1:g’(1)= -3/(4*1) +1/1 -1/4 = -3/4 +1 -1/4 = (-3/4 -1/4) +1 = -1 +1 =0.At x=e:g’(e)= -3/(4e²) +1/e -1/4. Let's compute this:-3/(4e²) +1/e -1/4 ≈ -3/(4*7.389) +0.3679 -0.25 ≈ -0.101 +0.3679 -0.25 ≈0.0169, which is positive.At x=2:g’(2)= -3/(4*4) +1/2 -1/4 = -3/16 +1/2 -1/4 = (-3/16) + (8/16 -4/16)= (-3 +4)/16=1/16 >0.So when a=4/3, the derivative at x=1 is zero, and positive elsewhere on [1,e]. Therefore, g(x) is increasing on [1,e]. If a>4/3, then the derivative becomes even more positive. If a<4/3, then at x=1, g’(1)= -1/(a*1) +1 -1/4= -1/a +3/4. For a <4/3, -1/a +3/4 would be less than -3/4 +3/4=0, hence negative. So the derivative at x=1 would be negative if a <4/3, which would mean g(x) is decreasing there, contradicting the requirement that g is increasing on the entire interval. Therefore, the minimal a is 4/3. So the range is [4/3, ∞).Therefore, part II answer is a ≥4/3.To recap:Part I: When a=1/2, f(x) has maximum 0 at x=1 and minimum ln2 -1 at x=2.Part II: The range of a is [4/3, ∞).Final Answer(I) The maximum value is boxed{0} and the minimum value is boxed{ln 2 - 1}.(II) The range of values for ( a ) is boxed{[frac{4}{3}, +infty)}.</think>Given the function ( f(x) = frac{1-x}{ax} + ln x ):(I) When ( a = frac{1}{2} ), we need to find the maximum and minimum values of ( f(x) ) on the interval ([1, e]). Substituting ( a = frac{1}{2} ) into the function, we get:[f(x) = frac{2(1 - x)}{x} + ln x = frac{2}{x} - 2 + ln x]To find the critical points, we compute the derivative:[f'(x) = -frac{2}{x^2} + frac{1}{x} = frac{x - 2}{x^2}]Setting ( f'(x) = 0 ) gives ( x = 2 ). Evaluating ( f(x) ) at the critical point and the endpoints:[f(1) = 0, quad f(2) = ln 2 - 1, quad f(e) = frac{2}{e} - 1]Comparing these values, the maximum is at ( x = 1 ) with value ( 0 ), and the minimum is at ( x = 2 ) with value ( ln 2 - 1 ).(II) To determine the range of values for ( a ) such that ( g(x) = f(x) - frac{1}{4}x ) is increasing on ([1, e]), we need ( g'(x) geq 0 ). The function ( g(x) ) is:[g(x) = frac{1 - x}{ax} + ln x - frac{1}{4}x]Computing the derivative:[g'(x) = -frac{1}{ax^2} + frac{1}{x} - frac{1}{4}]Rearranging and simplifying the inequality ( g'(x) geq 0 ):[a geq frac{1}{x - frac{x^2}{4}}]We need the maximum value of ( frac{4}{x(4 - x)} ) on ([1, e]). Evaluating at critical points and endpoints, the maximum value is ( frac{4}{3} ) at ( x = 1 ). Therefore, ( a geq frac{4}{3} ). Final Answer(I) The maximum value is boxed{0} and the minimum value is boxed{ln 2 - 1}.(II) The range of values for ( a ) is boxed{[frac{4}{3}, +infty)}.,<think>Okay, so I have this problem here with two parts. Let me start with part (I). The function is given as f(x) = (1 - x)/(a x) + ln x, and when a is 1/2, I need to find the maximum and minimum values of f(x) on the interval [1, e]. Alright, let's break this down.First, when a is 1/2, substitute that into the function. So f(x) becomes (1 - x)/( (1/2) x ) + ln x. Let me simplify that. The denominator is (1/2)x, so dividing by that is the same as multiplying by 2. So, (1 - x) * 2 / x + ln x. Which simplifies to 2(1 - x)/x + ln x. Let's write that out: 2*(1/x - 1) + ln x. So f(x) = 2/x - 2 + ln x. Got that.Now, to find the maximum and minimum on [1, e], I need to check the critical points and the endpoints. Critical points occur where the derivative is zero or undefined. Since the function involves 2/x and ln x, which are both defined and differentiable on (0, ∞), so on [1, e] there's no issue. So first, find f'(x).f'(x) is derivative of 2/x, which is -2/x², plus derivative of -2, which is 0, plus derivative of ln x, which is 1/x. So f'(x) = -2/x² + 1/x. Let me write that as ( -2 + x ) / x². Because combining the terms: (-2/x²) + (1/x) = (-2 + x)/x². Wait, let me check that again. If we have -2/x² + 1/x, to combine over a common denominator of x², it would be -2 + x. So yes, f'(x) = (x - 2)/x².Alright, so critical points occur when f'(x) = 0. So set (x - 2)/x² = 0. That happens when numerator is zero, so x = 2. So x=2 is the critical point. Now, check if x=2 is in the interval [1, e]. Since e is approximately 2.718, so 2 is between 1 and e. So yes, x=2 is in the interval. Therefore, we need to evaluate f(x) at x=1, x=2, and x=e.So compute f(1), f(2), f(e).First, f(1) = 2/1 - 2 + ln 1 = 2 - 2 + 0 = 0.f(2): 2/2 - 2 + ln 2 = 1 - 2 + ln 2 = -1 + ln 2. Since ln 2 is approximately 0.693, so that's about -0.307.f(e): 2/e - 2 + ln e. ln e is 1, so 2/e - 2 + 1 = 2/e -1. Since e is about 2.718, 2/e is approx 0.735, so 0.735 -1 = -0.265.So now compare these three values: f(1)=0, f(2)≈-0.307, f(e)≈-0.265.Therefore, the maximum value is 0 at x=1, and the minimum value is approximately -0.307 at x=2. But since we need exact values, not approximate. Let's write them exactly.f(2) is -1 + ln 2. f(e) is 2/e -1 +1? Wait, hold on. Wait, f(e) is 2/e - 2 + 1? Wait, wait. Wait, let's double-check that computation.Wait, f(x) = 2/x -2 + ln x. So f(e) = 2/e -2 + ln e. ln e is 1. So 2/e -2 +1 = 2/e -1. So yes, that's correct. So exact value is 2/e -1. Similarly, f(2) is -1 + ln 2.So comparing f(1)=0, f(2)= -1 + ln2, f(e)=2/e -1.Which one is smaller between f(2) and f(e)? Let's compute the exact difference. Let's see: which is more negative, -1 + ln2 or 2/e -1.Compute the difference between f(2) and f(e): (-1 + ln2) - (2/e -1) = ln2 - 2/e. Since ln2 ≈0.693 and 2/e≈0.735, so ln2 - 2/e≈-0.042. So f(2) is less than f(e). Therefore, the minimum is at x=2, which is -1 + ln2, and maximum at x=1, which is 0.So for part (I), the maximum is 0 at x=1, and minimum is -1 + ln2 at x=2.Now moving on to part (II). The function g(x) = f(x) - (1/4)x is increasing on [1,e]. Need to find the range of positive real numbers a for which this is true.First, since g is increasing on [1,e], its derivative g’(x) must be ≥0 for all x in [1,e].So let's find g’(x). Since g(x) = f(x) - (1/4)x, then g’(x) = f’(x) - 1/4.We already have f’(x). From the original function f(x) = (1 -x)/(a x) + ln x. Let's compute f’(x). First term: derivative of (1 -x)/(a x). Let me write that as (1/(a x)) - (x)/(a x) = (1/(a x)) - 1/a. Wait, but actually, let's compute derivative properly.Let f(x) = (1 - x)/(a x) + ln x. Let's compute f’(x):First term: d/dx [ (1 - x)/(a x) ].Let me write that as (1 - x)/ (a x) = (1/(a x) - x/(a x)) = (1/(a x) - 1/a). Then derivative is derivative of 1/(a x) is -1/(a x²), derivative of -1/a is 0. So that part's derivative is -1/(a x²). Then the derivative of ln x is 1/x. So f’(x) = -1/(a x²) + 1/x.Therefore, g’(x) = f’(x) - 1/4 = [ -1/(a x²) + 1/x ] - 1/4.So, g’(x) = -1/(a x²) + 1/x - 1/4.We need g’(x) ≥ 0 for all x in [1,e]. So, set up the inequality:-1/(a x²) + 1/x - 1/4 ≥ 0, for all x in [1,e].We need to find the values of a > 0 such that this inequality holds for all x in [1,e].Let's rearrange the inequality:-1/(a x²) ≥ -1/x + 1/4Multiply both sides by -1, which reverses the inequality:1/(a x²) ≤ 1/x - 1/4Note that 1/x - 1/4 must be positive for this inequality to make sense because the left side is positive (since a and x² are positive). Therefore, we also require that 1/x - 1/4 > 0, which implies 1/x > 1/4 => x < 4. Since x is in [1,e], and e ≈2.718 <4, so 1/x -1/4 is positive for all x in [1,e]. Therefore, the inequality is valid.So, from 1/(a x²) ≤ 1/x - 1/4,we can solve for a:Multiply both sides by a x² (positive, so inequality direction remains):1 ≤ a x² (1/x - 1/4 )Simplify the right-hand side:a x² (1/x - 1/4 ) = a x² ( (1/x) - (1/4) ) = a (x²*(1/x) - x²*(1/4)) = a (x - x²/4 )Therefore, 1 ≤ a (x - x²/4 )Then, solving for a:a ≥ 1 / (x - x²/4 )So, since this must hold for all x in [1,e], the maximum of 1/(x - x²/4 ) on [1,e] will be the lower bound for a. Therefore, a must be greater or equal to the maximum value of 1/(x - x²/4 ) over x in [1,e].Therefore, we need to find the maximum of h(x) = 1/(x - (x²)/4 ) on [1,e], and set a ≥ that maximum.So first, let's analyze h(x) = 1/(x - x²/4 ). Let's write the denominator as x - x²/4 = x(1 - x/4 ). Since x is in [1,e], and e ≈2.718 <4, so 1 - x/4 is positive (since x <4), so denominator is positive. Therefore, h(x) is positive.We need to find the maximum of h(x) on [1,e]. To find the maximum, we can take the derivative of h(x) and find its critical points.Let me compute h(x) = 1/(x - x²/4 ) = [x(1 - x/4 )]^{-1} = [x - x²/4 ]^{-1}Let’s compute h’(x):h’(x) = - [x - x²/4 ]^{-2} * (1 - (2x)/4 )Simplify:h’(x) = - [x - x²/4 ]^{-2} * (1 - x/2 )Set derivative to zero:- [x - x²/4 ]^{-2} * (1 - x/2 ) =0Since [x - x²/4 ]^{-2} is never zero, the critical points occur when 1 - x/2 =0 => x=2.So the critical point is at x=2. Now, we need to evaluate h(x) at x=1, x=2, x=e, to find which gives the maximum.Compute h(1):h(1) = 1/(1 - 1²/4 ) = 1/(1 - 1/4 ) = 1/(3/4 ) = 4/3 ≈1.333.h(2):h(2) = 1/(2 - (2²)/4 ) = 1/(2 -4/4 )=1/(2 -1)=1/1=1.h(e):h(e)=1/(e - e²/4 ). Let's compute e - e²/4. Since e≈2.718, e²≈7.389, so e²/4≈1.847. Therefore, e - e²/4 ≈2.718 -1.847≈0.871. So h(e)≈1/0.871≈1.148.Therefore, h(1)=4/3≈1.333, h(2)=1, h(e)≈1.148. So the maximum of h(x) on [1,e] is at x=1, which is 4/3.Therefore, the maximum of h(x) is 4/3 at x=1, so a must be greater or equal to 4/3.But wait, let me confirm if h(x) indeed attains its maximum at x=1. Since the derivative h’(x) is negative when x >2 (since 1 -x/2 is negative), but our interval is up to e≈2.718. So between x=2 and x=e, h’(x) is negative, so h(x) is decreasing. From x=1 to x=2, since the critical point is at x=2, but in the interval [1,2], h’(x) is positive or negative? Let's check.h’(x) = - [denominator]^{-2} * (1 -x/2 ). The sign of h’(x) is determined by the sign of -(1 -x/2 ). So:If x <2, then 1 -x/2 >0, so h’(x) = - positive = negative.Wait, wait, hold on. Wait, h’(x) is equal to - [denominator]^{-2} * (1 - x/2 )Denominator is positive, so denominator^{-2} is positive. Therefore, the sign is determined by - (1 - x/2 ). So:If 1 - x/2 >0 (i.e., x <2 ), then h’(x) is negative.If 1 -x/2 <0 (i.e., x>2 ), then h’(x) is positive.But in our interval [1,e], which is up to e≈2.718. So for x in [1,2), h’(x) is negative; for x in (2,e], h’(x) is positive.Wait, that can't be. Wait, if x <2, 1 -x/2 >0, so h’(x)= - positive * positive = negative. If x>2, 1 -x/2 <0, so h’(x)= - positive * negative = positive. Therefore, h(x) is decreasing on [1,2) and increasing on (2,e]. Therefore, the minimum of h(x) occurs at x=2, and the maximum could be at the endpoints x=1 or x=e.Wait, but earlier we computed h(1)=4/3≈1.333, h(e)≈1.148, so h(1) is larger. Therefore, the maximum of h(x) on [1,e] is at x=1, and the minimum at x=2. So h(x) decreases from x=1 to x=2, then increases from x=2 to x=e, but since h(e)≈1.148 which is less than h(1)=1.333, so maximum is at x=1.Therefore, the maximum of h(x) on [1,e] is h(1)=4/3. Therefore, a must be ≥4/3.But let me double-check this conclusion. So, since h(x) achieves its maximum at x=1, then to satisfy 1/(x -x²/4 ) ≤a for all x in [1,e], the smallest possible a is the maximum of 1/(x -x²/4 ), which is 4/3. Therefore, the range of a is [4/3, ∞).Therefore, the answer for part (II) is a ≥4/3.Wait, but let me verify with an example. Suppose a=4/3. Then, check if g’(x) ≥0 for all x in [1,e].g’(x) = -1/(a x²) +1/x -1/4. Substitute a=4/3:g’(x) = -1/( (4/3)x² ) +1/x -1/4 = -3/(4x²) +1/x -1/4.We need to check if this is ≥0 for all x in [1,e].Take x=1:g’(1)= -3/4 +1 -1/4= (-3/4 -1/4) +1= -1 +1=0.At x=2:g’(2)= -3/(4*4) +1/2 -1/4= -3/16 + 1/2 -1/4= (-3/16) + (8/16 -4/16)= (-3 +4)/16=1/16 >0.At x=e:Compute g’(e)= -3/(4e²) +1/e -1/4≈ -3/(4*7.389) +0.368 -0.25≈ -0.101 +0.368 -0.25≈0.017>0.So at x=1, g’(1)=0, and at other points it's positive, so a=4/3 makes g’(x)≥0 on [1,e]. If a is larger than 4/3, then -1/(a x²) becomes less negative, so the entire expression g’(x) becomes larger, so the inequality still holds. If a is smaller than 4/3, then at x=1, g’(1)= -1/(a*1) +1 -1/4= -1/a +3/4. If a <4/3, then 1/a >3/4, so -1/a +3/4 becomes negative, which would violate the condition. Therefore, the minimum a is 4/3.Therefore, the answer for part (II) is a ≥4/3, so the range is [4/3, ∞).Final Answer(I) The maximum value is boxed{0} and the minimum value is boxed{-1 + ln 2}.(II) The range of values for ( a ) is boxed{[frac{4}{3}, infty)}.</think>Given the function ( f(x) = frac{1 - x}{ax} + ln x ):(I) When ( a = frac{1}{2} ), we need to find the maximum and minimum values of ( f(x) ) on the interval ([1, e]).1. Substitute ( a = frac{1}{2} ) into the function: [ f(x) = frac{1 - x}{frac{1}{2}x} + ln x = frac{2(1 - x)}{x} + ln x = frac{2}{x} - 2 + ln x ]2. Compute the derivative ( f'(x) ): [ f'(x) = -frac{2}{x^2} + frac{1}{x} = frac{x - 2}{x^2} ]3. Find critical points by setting ( f'(x) = 0 ): [ frac{x - 2}{x^2} = 0 implies x = 2 ]4. Evaluate ( f(x) ) at the critical point ( x = 2 ) and the endpoints ( x = 1 ) and ( x = e ): [ f(1) = 0, quad f(2) = -1 + ln 2, quad f(e) = frac{2}{e} - 1 ]5. Compare the values: - Maximum value: ( 0 ) at ( x = 1 ) - Minimum value: ( -1 + ln 2 ) at ( x = 2 )(II) If the function ( g(x) = f(x) - frac{1}{4}x ) is increasing on ([1, e]), find the range of values for ( a ).1. Compute the derivative ( g'(x) ): [ g'(x) = f'(x) - frac{1}{4} = left( -frac{1}{a x^2} + frac{1}{x} right) - frac{1}{4} ]2. Set ( g'(x) geq 0 ): [ -frac{1}{a x^2} + frac{1}{x} - frac{1}{4} geq 0 ]3. Rearrange and solve for ( a ): [ frac{1}{a x^2} leq frac{1}{x} - frac{1}{4} implies a geq frac{1}{x - frac{x^2}{4}} ]4. Find the maximum value of ( frac{1}{x - frac{x^2}{4}} ) on ([1, e]): - Critical point at ( x = 2 ) - Evaluate at ( x = 1 ), ( x = 2 ), and ( x = e ): [ frac{1}{1 - frac{1}{4}} = frac{4}{3}, quad frac{1}{2 - 1} = 1, quad frac{1}{e - frac{e^2}{4}} approx 1.148 ] - Maximum value is ( frac{4}{3} ) at ( x = 1 )Final answers:(I) The maximum value is (boxed{0}) and the minimum value is (boxed{-1 + ln 2}).(II) The range of values for ( a ) is (boxed{[frac{4}{3}, infty)}).]

🔑:(I) When a=frac{1}{2}, we have f'(x)=frac{x-2}{x^{2}}(x > 0).- When xin[1,2), f'(x) 0, so f(x) is increasing on (2,e].Thus, f(x) has a unique minimum value on the interval [1,e]. The minimum value is f(x)_{text{min}}=f(x)_{text{extreme}}=f(2)=ln 2-1.Also, since f(1)=0 and f(e)=frac{2-e}{e} 0). Let h(x)=-ax^{2}+4ax-4. We need h(x)geqslant 0 to hold true on the interval [1,e]. Since a > 0, the axis of symmetry for the graph of h(x) is x=2. Therefore, we only need h(1)=3a-4geqslant 0, which implies ageqslant frac{4}{3}. So, the range of values for a is boxed{[frac{4}{3},+infty)}.

🧠:[<think>Okay, so I need to find the equation of a circle that passes through three points: A(4, 1), B(6, -3), and C(-3, 0). Hmm, let's remember how to find the equation of a circle given three points. I think the general equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. Since we have three points, maybe I can set up three equations and solve for h, k, and r. But solving three equations might be a bit tedious. Wait, there's another method using the perpendicular bisectors of two chords to find the center. Since the perpendicular bisector of any chord passes through the center of the circle, maybe that's a better approach. Let me try that.First, let's take two chords. Let's take chord AB and chord AC. Wait, but chord AB is between points A(4,1) and B(6,-3), and chord AC is between A(4,1) and C(-3,0). Alternatively, maybe chord BC would be better. Let me just pick two chords. Let's start with AB and BC.For chord AB: points A(4,1) and B(6,-3). The midpoint of AB would be the average of the coordinates. So, midpoint M1 is ((4 + 6)/2, (1 + (-3))/2) = (5, -1). The slope of AB is ( -3 - 1 ) / (6 - 4 ) = (-4)/2 = -2. Therefore, the perpendicular bisector will have a slope that's the negative reciprocal, so 1/2. So the equation of the perpendicular bisector of AB is y - (-1) = (1/2)(x - 5). Simplifying, that's y + 1 = (1/2)x - 5/2, so y = (1/2)x - 5/2 - 1 = (1/2)x - 7/2.Now, let's do the same for chord BC: points B(6,-3) and C(-3,0). The midpoint M2 is ((6 + (-3))/2, (-3 + 0)/2) = (3/2, -3/2). The slope of BC is (0 - (-3))/(-3 - 6) = 3/(-9) = -1/3. Therefore, the perpendicular bisector will have a slope of 3. So the equation of the perpendicular bisector of BC is y - (-3/2) = 3(x - 3/2). Simplifying, y + 3/2 = 3x - 9/2. Subtract 3/2: y = 3x - 9/2 - 3/2 = 3x - 12/2 = 3x - 6.Now, the intersection of these two perpendicular bisectors should give the center of the circle. So we have two equations:1. y = (1/2)x - 7/22. y = 3x - 6Set them equal to each other:(1/2)x - 7/2 = 3x - 6Multiply both sides by 2 to eliminate denominators:x - 7 = 6x - 12Subtract x from both sides:-7 = 5x - 12Add 12 to both sides:5 = 5xSo x = 1. Then plug back into one of the equations to find y. Let's use equation 2:y = 3(1) - 6 = 3 - 6 = -3.Wait, so the center is (1, -3)? Let me check that with equation 1:y = (1/2)(1) - 7/2 = 1/2 - 7/2 = (-6)/2 = -3. Okay, that's consistent. So center at (1, -3).Now, let's find the radius. We can use the distance from the center to any of the three points. Let's use point A(4,1). Distance formula:r = sqrt[(4 - 1)^2 + (1 - (-3))^2] = sqrt[(3)^2 + (4)^2] = sqrt[9 + 16] = sqrt[25] = 5.So the equation of the circle should be (x - 1)^2 + (y + 3)^2 = 25. Let me check if this works for the other points.Testing point B(6, -3):(6 - 1)^2 + (-3 + 3)^2 = 5^2 + 0^2 = 25 + 0 = 25. Correct.Testing point C(-3, 0):(-3 - 1)^2 + (0 + 3)^2 = (-4)^2 + 3^2 = 16 + 9 = 25. Correct.So all three points satisfy the equation. Therefore, the equation of the circle is (x - 1)^2 + (y + 3)^2 = 25.Wait, but let me just make sure I didn't make any calculation errors. Let me recalculate the midpoint of AB. (4+6)/2 = 10/2 = 5, (1 + (-3))/2 = (-2)/2 = -1. That's correct. The slope of AB was (-4)/2 = -2, so perpendicular slope 1/2. Equation: y + 1 = 1/2(x - 5). Which simplifies to y = (1/2)x - 5/2 -1 = (1/2)x - 7/2. Correct.For chord BC: midpoint (6 + (-3))/2 = 3/2, (-3 + 0)/2 = -3/2. Correct. Slope of BC: (0 - (-3))/(-3 -6) = 3/-9 = -1/3. Perpendicular slope 3. Equation: y + 3/2 = 3(x - 3/2). Then y = 3x - 9/2 - 3/2 = 3x - 12/2 = 3x -6. Correct.Solving the two equations: (1/2)x -7/2 = 3x -6. Multiply by 2: x -7 = 6x -12. Then -7 +12 = 6x -x. 5 =5x. x=1. Then y=3(1)-6=-3. Correct.Radius from (1, -3) to (4,1): delta x=3, delta y=4. sqrt(9+16)=5. All points check out. So equation is correct.Final AnswerThe equation of circle C is boxed{(x - 1)^2 + (y + 3)^2 = 25}.</think>To find the equation of the circle passing through points A(4, 1), B(6, -3), and C(-3, 0), we use the perpendicular bisectors of two chords to determine the center of the circle.1. Perpendicular Bisector of Chord AB: - Midpoint of AB: (left( frac{4+6}{2}, frac{1 + (-3)}{2} right) = (5, -1)) - Slope of AB: (frac{-3 - 1}{6 - 4} = -2) - Perpendicular slope: (frac{1}{2}) - Equation: (y + 1 = frac{1}{2}(x - 5)) simplifies to (y = frac{1}{2}x - frac{7}{2})2. Perpendicular Bisector of Chord BC: - Midpoint of BC: (left( frac{6 + (-3)}{2}, frac{-3 + 0}{2} right) = left( frac{3}{2}, -frac{3}{2} right)) - Slope of BC: (frac{0 - (-3)}{-3 - 6} = -frac{1}{3}) - Perpendicular slope: (3) - Equation: (y + frac{3}{2} = 3left(x - frac{3}{2}right)) simplifies to (y = 3x - 6)3. Finding the Intersection of Perpendicular Bisectors: - Solve the system: [ begin{cases} y = frac{1}{2}x - frac{7}{2} y = 3x - 6 end{cases} ] - Setting the equations equal: (frac{1}{2}x - frac{7}{2} = 3x - 6) - Solving for (x): (x = 1) - Substituting (x = 1) into (y = 3x - 6): (y = -3) - Center: ((1, -3))4. Calculating the Radius: - Using point A(4, 1): [ r = sqrt{(4 - 1)^2 + (1 - (-3))^2} = sqrt{3^2 + 4^2} = sqrt{25} = 5 ]5. Equation of the Circle: - ((x - 1)^2 + (y + 3)^2 = 25)All points A, B, and C satisfy this equation. Therefore, the equation of circle C is (boxed{(x - 1)^2 + (y + 3)^2 = 25}).,<think>Okay, so I need to find the equation of a circle that passes through three points: A(4, 1), B(6, -3), and C(-3, 0). Hmm, let me remember how to do this. I think the general equation of a circle is something like (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. Since the circle passes through these three points, I can plug each of them into this equation and get three equations with variables h, k, and r. Then I can solve those equations to find h, k, and r. Let me start by writing down the three equations. For point A(4,1):(4 - h)^2 + (1 - k)^2 = r^2.For point B(6, -3):(6 - h)^2 + (-3 - k)^2 = r^2.For point C(-3, 0):(-3 - h)^2 + (0 - k)^2 = r^2.So now I have three equations:1) (4 - h)^2 + (1 - k)^2 = r^2.2) (6 - h)^2 + (-3 - k)^2 = r^2.3) (-3 - h)^2 + (0 - k)^2 = r^2.Since all three equal r^2, I can set them equal to each other. Let me subtract equation 1 from equation 2 to eliminate r^2. That might give me a linear equation in h and k. So, equation 2 minus equation 1:[(6 - h)^2 + (-3 - k)^2] - [(4 - h)^2 + (1 - k)^2] = 0.Let me expand these squares. First, (6 - h)^2 is (36 - 12h + h^2), (-3 - k)^2 is (9 + 6k + k^2), (4 - h)^2 is (16 - 8h + h^2), and (1 - k)^2 is (1 - 2k + k^2).Substituting back:[36 - 12h + h^2 + 9 + 6k + k^2] - [16 - 8h + h^2 + 1 - 2k + k^2] = 0.Simplify the terms inside the brackets:First part: 36 + 9 = 45, -12h + 6k, h^2 + k^2.Second part: 16 + 1 = 17, -8h -2k, h^2 + k^2.So subtracting the second part from the first:45 - 17 + (-12h + 8h) + (6k + 2k) + (h^2 + k^2 - h^2 - k^2) = 0.Simplify:28 - 4h + 8k = 0.So, -4h + 8k = -28.Divide both sides by -4:h - 2k = 7. Let me call this equation 4.Okay, now let's do the same with equations 1 and 3. Subtract equation 1 from equation 3:[(-3 - h)^2 + (0 - k)^2] - [(4 - h)^2 + (1 - k)^2] = 0.Expanding each term:(-3 - h)^2 = 9 + 6h + h^2,(0 - k)^2 = k^2,(4 - h)^2 = 16 - 8h + h^2,(1 - k)^2 = 1 - 2k + k^2.Substituting back:[9 + 6h + h^2 + k^2] - [16 - 8h + h^2 + 1 - 2k + k^2] = 0.Simplify:First part: 9 + 6h + h^2 + k^2,Second part: 16 + 1 = 17, -8h -2k, h^2 + k^2.Subtracting second part from first:9 - 17 + 6h + 8h + 0 + 2k + (h^2 + k^2 - h^2 - k^2) = 0.Simplify:-8 + 14h + 2k = 0.So, 14h + 2k = 8.Divide both sides by 2:7h + k = 4. Let's call this equation 5.Now we have two equations:Equation 4: h - 2k = 7.Equation 5: 7h + k = 4.We can solve this system of equations. Let's use substitution or elimination. Maybe elimination.Multiply equation 5 by 2: 14h + 2k = 8.Now add equation 4 to this:14h + 2k + h - 2k = 8 + 7.15h = 15.So h = 1.Now substitute h = 1 into equation 4:1 - 2k = 7.-2k = 6.k = -3.So the center of the circle is (h, k) = (1, -3). Now we need to find the radius r. Let's plug one of the points into the equation. Let's use point A(4,1):(4 - 1)^2 + (1 - (-3))^2 = r^2.So (3)^2 + (4)^2 = r^2.9 + 16 = 25 = r^2.So r = 5.Therefore, the equation of the circle is (x - 1)^2 + (y + 3)^2 = 25.But let me verify with another point to be sure. Let's check point B(6, -3):(6 -1)^2 + (-3 + 3)^2 = 5^2 + 0^2 = 25 + 0 = 25. Correct.Check point C(-3,0):(-3 -1)^2 + (0 +3)^2 = (-4)^2 + 3^2 = 16 + 9 = 25. Correct. Okay, that works.Wait, but let me just check my algebra again because sometimes signs can be tricky. When I expanded (-3 - h)^2, that should be ( -3 - h )^2 = ( - (3 + h) )^2 = (3 + h)^2 = 9 + 6h + h^2. Wait, but is that correct? Wait, (-3 - h)^2 is the same as (h + 3)^2, which is h^2 + 6h + 9. Yes, that's right. So that part was correct.Similarly, (0 - k)^2 is k^2. Correct. Then when subtracting the second part, which is (4 - h)^2 + (1 - k)^2 = (16 - 8h + h^2) + (1 - 2k + k^2). So that's 16 + 1 =17, -8h -2k, and h^2 + k^2. Then subtracting that gives 9 -17 = -8, 6h - (-8h) = 14h, k^2 - k^2 = 0, and then +2k. So yes, -8 +14h +2k=0. Then 14h +2k=8, divide by 2:7h +k=4. Correct.Then solving equations 4 and 5: h=1, k=-3. That seems right. Then radius squared is 25. All points check out. So the equation is (x -1)^2 + (y +3)^2 =25. So I think that's the answer.Final AnswerThe equation of circle C is boxed{(x - 1)^2 + (y + 3)^2 = 25}.</think>Given three points A(4, 1), B(6, -3), and C(-3, 0) on a circle, we need to find the equation of the circle. The general equation of a circle is ((x - h)^2 + (y - k)^2 = r^2), where ((h, k)) is the center and (r) is the radius.We start by setting up three equations based on the given points:1. For point A(4, 1): [ (4 - h)^2 + (1 - k)^2 = r^2 ]2. For point B(6, -3): [ (6 - h)^2 + (-3 - k)^2 = r^2 ]3. For point C(-3, 0): [ (-3 - h)^2 + (0 - k)^2 = r^2 ]Subtracting the first equation from the second to eliminate (r^2):[[(6 - h)^2 + (-3 - k)^2] - [(4 - h)^2 + (1 - k)^2] = 0]Expanding and simplifying:[45 - 12h + 6k - (17 - 8h - 2k) = 0 implies 28 - 4h + 8k = 0 implies h - 2k = 7 quad text{(Equation 4)}]Subtracting the first equation from the third to eliminate (r^2):[[(-3 - h)^2 + k^2] - [(4 - h)^2 + (1 - k)^2] = 0]Expanding and simplifying:[9 + 6h - (17 - 8h - 2k) = 0 implies -8 + 14h + 2k = 0 implies 7h + k = 4 quad text{(Equation 5)}]Solving the system of equations (Equation 4 and Equation 5):1. (h - 2k = 7)2. (7h + k = 4)Multiply Equation 5 by 2:[14h + 2k = 8]Add to Equation 4:[15h = 15 implies h = 1]Substitute (h = 1) into Equation 4:[1 - 2k = 7 implies -2k = 6 implies k = -3]The center of the circle is ((1, -3)). Using point A(4, 1) to find (r):[(4 - 1)^2 + (1 + 3)^2 = r^2 implies 3^2 + 4^2 = r^2 implies 9 + 16 = 25 implies r = 5]Verification with points B and C confirms the radius. The equation of the circle is:[boxed{(x - 1)^2 + (y + 3)^2 = 25}]]

🔑:Let's assume the general equation of circle C is x^2+y^2+Dx+Ey+F=0.Since points A(4, 1), B(6, -3), and C(-3, 0) lie on the circle, we have the system of equations:begin{cases} 17+4D+E+F=0 45+6D-3E+F=0 9-3D+F=0end{cases}Solving this system, we find D=-2, E=6, and F=-15.Therefore, the equation of circle C is x^2+y^2-2x+6y-15=0.Hence, the answer is boxed{x^2+y^2-2x+6y-15=0}.By setting up the general equation of a circle and substituting the coordinates of points A(4, 1), B(6, -3), and C(-3, 0), we can solve for the values of D, E, and F to find the equation of the circle. This problem mainly tests the ability to find the equation of a circle using the method of undetermined coefficients, assessing the computational skills of the students. It is considered a medium-level question.

🧠:[<think>Okay, so I need to express the polynomial ( x^4 + x^3 + x^2 + x + 1 ) as the difference of squares of two real-coefficient polynomials of different degrees. Hmm, difference of squares... That means I need to find two polynomials, say ( A(x) ) and ( B(x) ), such that ( A(x)^2 - B(x)^2 = x^4 + x^3 + x^2 + x + 1 ). Also, they must have different degrees and real coefficients. Let's think about how to approach this.First, since the given polynomial is a quartic (degree 4), the highest degree term in the difference of squares must come from the square of the highest degree term in ( A(x) ) or ( B(x) ). Let's assume that ( A(x) ) is of higher degree than ( B(x) ). So if ( A(x) ) is, say, quadratic (degree 2), then ( A(x)^2 ) would be quartic, which matches the degree of the given polynomial. Then ( B(x) ) would need to be of lower degree, maybe linear or quadratic. Wait, but the problem says they have different degrees, so if ( A(x) ) is quadratic, then ( B(x) ) can't be quadratic. So maybe ( B(x) ) is linear or cubic? Wait, but ( A(x)^2 - B(x)^2 ) is quartic. If ( A(x) ) is quadratic (degree 2), then ( A(x)^2 ) is degree 4. If ( B(x) ) is linear (degree 1), then ( B(x)^2 ) is degree 2. Then the difference would be degree 4 minus degree 2, which is still degree 4. That works. Alternatively, if ( B(x) ) is cubic, then ( B(x)^2 ) would be degree 6, which is higher than 4, so that would make the difference have degree 6, which is not desired. So probably ( A(x) ) is quadratic and ( B(x) ) is linear. Let me try that.Let me denote ( A(x) = ax^2 + bx + c ) and ( B(x) = dx + e ). Then compute ( A(x)^2 - B(x)^2 ):( (ax^2 + bx + c)^2 - (dx + e)^2 )Expanding both squares:( a^2x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2 - (d^2x^2 + 2dex + e^2) )Combine like terms:( a^2x^4 + 2abx^3 + (2ac + b^2 - d^2)x^2 + (2bc - 2de)x + (c^2 - e^2) )We need this to equal ( x^4 + x^3 + x^2 + x + 1 ). So we can set up equations by equating the coefficients:1. Coefficient of ( x^4 ): ( a^2 = 1 )2. Coefficient of ( x^3 ): ( 2ab = 1 )3. Coefficient of ( x^2 ): ( 2ac + b^2 - d^2 = 1 )4. Coefficient of ( x ): ( 2bc - 2de = 1 )5. Constant term: ( c^2 - e^2 = 1 )Okay, now we have a system of equations. Let's try solving them step by step.From equation 1: ( a^2 = 1 ). So ( a = 1 ) or ( a = -1 ). Let's try ( a = 1 ) first. If that doesn't work, we can try ( a = -1 ).From equation 2: ( 2ab = 1 ). Since ( a = 1 ), then ( 2b = 1 implies b = 1/2 ).Now equation 3: ( 2ac + b^2 - d^2 = 1 ). Substitute ( a = 1 ), ( c ) is still unknown, ( b = 1/2 ). So:( 2*1*c + (1/2)^2 - d^2 = 1 implies 2c + 1/4 - d^2 = 1 implies 2c - d^2 = 1 - 1/4 = 3/4 ). So equation 3 becomes:( 2c - d^2 = 3/4 ). Let's note that.Equation 4: ( 2bc - 2de = 1 ). Substituting ( b = 1/2 ), we get:( 2*(1/2)*c - 2de = 1 implies c - 2de = 1 ).Equation 5: ( c^2 - e^2 = 1 ).So now we have three equations:3. ( 2c - d^2 = 3/4 )4. ( c - 2de = 1 )5. ( c^2 - e^2 = 1 )We need to solve for c, d, e.This seems a bit complicated. Let's see if we can express variables in terms of others.From equation 4: ( c = 1 + 2de ). Let's substitute this into equation 3.Substitute c into equation 3:( 2*(1 + 2de) - d^2 = 3/4 implies 2 + 4de - d^2 = 3/4 implies 4de - d^2 = 3/4 - 2 = -5/4 ).So:( -d^2 + 4de = -5/4 )Multiply both sides by -1:( d^2 - 4de = 5/4 )Hmm, maybe factor this?Alternatively, let's see if we can express this in terms of e.Wait, equation 5 is ( c^2 - e^2 = 1 ), and we have c expressed as ( c = 1 + 2de ). So substitute c into equation 5:( (1 + 2de)^2 - e^2 = 1 )Expanding the square:( 1 + 4de + 4d^2e^2 - e^2 = 1 implies 4de + (4d^2e^2 - e^2) = 0 )Factor e^2 from the last two terms:( 4de + e^2(4d^2 - 1) = 0 )Let me factor e:( e(4d + e(4d^2 - 1)) = 0 )So either e = 0 or 4d + e(4d^2 - 1) = 0.If e = 0, then from equation 4: c = 1 + 0 = 1. Then from equation 3: 2*1 - d^2 = 3/4 => 2 - d^2 = 3/4 => d^2 = 2 - 3/4 = 5/4 => d = sqrt(5)/2 or -sqrt(5)/2. Then check equation 5: c^2 - e^2 = 1^2 - 0 = 1, which works. But let's check equation 4 with e = 0: c = 1, which is okay. Then equation 4: 1 - 0 = 1, which holds. So e = 0 is possible. Then we have:If e = 0, then c = 1, d = sqrt(5)/2 or -sqrt(5)/2.But let's check if this works for all equations. Then, let's check the coefficient of x in the original polynomial. From equation 4: 2bc - 2de = 1. If e = 0, then 2bc = 1. But 2bc = 2*(1/2)*1 = 1, which works. So that's okay.Wait, but let's verify the entire polynomial. Let me write out A(x) and B(x) with these values:If a = 1, b = 1/2, c = 1, d = sqrt(5)/2, e = 0.Then A(x) = x^2 + (1/2)x + 1B(x) = (sqrt(5)/2)x + 0 = (sqrt(5)/2)xCompute A(x)^2 - B(x)^2:A(x)^2 = (x^2 + (1/2)x + 1)^2 = x^4 + x^3 + (1/4)x^2 + 2x^2 + x + 1Wait, let me expand it step by step:(x^2 + (1/2)x + 1)(x^2 + (1/2)x + 1)First multiply x^2 by each term: x^2*x^2 = x^4, x^2*(1/2)x = (1/2)x^3, x^2*1 = x^2Then multiply (1/2)x by each term: (1/2)x*x^2 = (1/2)x^3, (1/2)x*(1/2)x = (1/4)x^2, (1/2)x*1 = (1/2)xThen multiply 1 by each term: 1*x^2 = x^2, 1*(1/2)x = (1/2)x, 1*1 = 1Now add all the terms:x^4 + (1/2x^3 + 1/2x^3) + (x^2 + 1/4x^2 + x^2) + (1/2x + 1/2x) + 1Simplify:x^4 + x^3 + ( (1 + 0.25 + 1) )x^2 + (1x) +1Which is:x^4 + x^3 + (2.25)x^2 + x + 1Now subtract B(x)^2:B(x)^2 = (sqrt(5)/2 x)^2 = (5/4)x^2So A(x)^2 - B(x)^2 = x^4 + x^3 + 2.25x^2 + x +1 - (5/4)x^2Since 5/4 is 1.25, so 2.25 -1.25 =1. So:x^4 + x^3 + x^2 + x +1Which is exactly the polynomial we wanted! So this works. Therefore, the polynomials are A(x) = x^2 + (1/2)x +1 and B(x) = (sqrt(5)/2)x.But let's check if they have different degrees. A(x) is degree 2, B(x) is degree 1. Yes, different degrees. And coefficients are real. So that's a valid solution.Wait, but in this case, A is quadratic and B is linear. So their squares are quartic and quadratic, respectively. The difference is quartic. So that works. Great!But let me check if there's another solution where e ≠0. Let's see.If e ≠0, then from equation above: 4d + e(4d^2 -1) =0. So 4d = -e(4d^2 -1). Then e = -4d / (4d^2 -1). Let's substitute this into equation from equation 3: d^2 -4de =5/4.Wait, let's recall:Earlier, after substituting c in terms of d and e, we had:d^2 -4de =5/4.But now, if e = -4d / (4d^2 -1), substitute that into this equation:d^2 -4d*(-4d / (4d^2 -1)) =5/4.Compute:d^2 + (16d^2)/(4d^2 -1) =5/4.Multiply both sides by 4d^2 -1 to eliminate denominator:d^2*(4d^2 -1) +16d^2 = (5/4)*(4d^2 -1)But this seems a bit complicated. Let me compute step by step.Left side: d^2*(4d^2 -1) +16d^2 =4d^4 -d^2 +16d^2 =4d^4 +15d^2Right side: (5/4)*(4d^2 -1) =5d^2 -5/4So set left = right:4d^4 +15d^2 =5d^2 -5/4Bring all terms to left:4d^4 +15d^2 -5d^2 +5/4=0Simplify:4d^4 +10d^2 +5/4=0Multiply both sides by 4 to eliminate fraction:16d^4 +40d^2 +5=0Let me set y =d^2. Then equation becomes:16y^2 +40y +5=0Quadratic in y: discriminant is 1600 - 320=1280Wait, discriminant D=40^2 -4*16*5=1600 -320=1280sqrt(1280)=sqrt(64*20)=8*sqrt(20)=8*2*sqrt(5)=16sqrt(5)Thus, roots are:y=(-40 ±16sqrt(5))/(2*16)= (-40 ±16sqrt(5))/32= (-5 ±2sqrt(5))/4But y =d^2, so d^2 must be real and non-negative. Let's compute (-5 +2sqrt(5))/4 and (-5 -2sqrt(5))/4.Compute sqrt(5) ≈2.236. So:First root: (-5 +2*2.236)/4≈(-5 +4.472)/4≈(-0.528)/4≈-0.132 <0Second root: (-5 -4.472)/4≈-9.472/4≈-2.368 <0Both roots negative. Thus, no real solutions. Hence, the only real solution is when e=0. Therefore, the previous solution is the only one.Therefore, the answer is A(x)=x^2 + (1/2)x +1 and B(x)= (sqrt(5)/2)x, so the difference of squares is (x^2 + (1/2)x +1)^2 - (sqrt(5)/2 x)^2.Alternatively, we can write the coefficients as fractions to make it cleaner. Let me check the values again:A(x) =x^2 + (1/2)x +1. So coefficients are 1, 1/2, 1.B(x) = (sqrt(5)/2)x. So coefficient is sqrt(5)/2.Alternatively, we can factor out 1/2 from A(x):Wait, A(x) =x^2 + (1/2)x +1. Let me see if we can write A(x) as (1/2)(2x^2 +x +2). Then A(x)^2 would be (1/2)^2*(2x^2 +x +2)^2 = (1/4)(4x^4 +4x^3 + ... ). Hmm, maybe not helpful. Alternatively, maybe write both A(x) and B(x) with denominator 2 to have integer coefficients?But the problem says real coefficients, so fractions are allowed. So perhaps the answer is as is.But let me check the problem statement again: "Express the polynomial... as the difference of squares of two real-coefficient polynomials of different degrees."So they just need to write the given polynomial as A(x)^2 - B(x)^2, where A and B have real coefficients and are of different degrees. Our solution has A(x) degree 2 and B(x) degree 1, which satisfies different degrees. The coefficients are real, so that works.Therefore, the final answer should be:( left(x^2 + frac{1}{2}x + 1right)^2 - left(frac{sqrt{5}}{2}xright)^2 )Let me verify once again by expanding this expression.First, expand ( left(x^2 + frac{1}{2}x + 1right)^2 ):= ( x^4 + x^3 + left(frac{1}{4} + 2right)x^2 + x + 1 )Wait, let's do it term by term:( (x^2)^2 = x^4 )( 2*x^2*(1/2x) = x^3 )( 2*x^2*1 = 2x^2 )( (1/2x)^2 = (1/4)x^2 )( 2*(1/2x)*1 = x )( 1^2 =1 )So combining all terms:x^4 + x^3 + (2x^2 + (1/4)x^2) + x +1= x^4 + x^3 + (9/4)x^2 +x +1Now subtract ( left(frac{sqrt{5}}{2}xright)^2 = (5/4)x^2 )So:x^4 + x^3 + (9/4)x^2 +x +1 - (5/4)x^2 = x^4 +x^3 + (4/4)x^2 +x +1 = x^4 +x^3 +x^2 +x +1Which matches exactly the given polynomial. Perfect.So this works. Therefore, the answer is the difference between the square of ( x^2 + frac{1}{2}x + 1 ) and the square of ( frac{sqrt{5}}{2}x ).Alternatively, if we want to avoid fractions in the coefficients, we might multiply both polynomials by 2 to eliminate denominators, but since the problem allows real coefficients, fractions are acceptable. However, let's check if there's another way with integer coefficients, but I think not necessarily. Because when we solved, we got fractions and sqrt(5), which is irrational, so it's required.Alternatively, if we take A(x) as quadratic and B(x) as linear, perhaps there's another set of coefficients? But from the system of equations, we saw that e=0 was the only real solution, so I think this is the unique solution in terms of A quadratic and B linear. Therefore, this is the answer.Final AnswerThe polynomial can be expressed as boxed{left(x^2 + frac{1}{2}x + 1right)^2 - left(frac{sqrt{5}}{2}xright)^2}.</think>To express the polynomial ( x^4 + x^3 + x^2 + x + 1 ) as the difference of squares of two real-coefficient polynomials of different degrees, we need to find polynomials ( A(x) ) and ( B(x) ) such that ( A(x)^2 - B(x)^2 = x^4 + x^3 + x^2 + x + 1 ).Assuming ( A(x) ) is a quadratic polynomial and ( B(x) ) is a linear polynomial, we start by setting:[ A(x) = ax^2 + bx + c ][ B(x) = dx + e ]Expanding ( A(x)^2 ) and ( B(x)^2 ):[ (ax^2 + bx + c)^2 = a^2x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2 ][ (dx + e)^2 = d^2x^2 + 2dex + e^2 ]Subtracting these expansions:[ a^2x^4 + 2abx^3 + (2ac + b^2 - d^2)x^2 + (2bc - 2de)x + (c^2 - e^2) ]Equating coefficients with ( x^4 + x^3 + x^2 + x + 1 ):1. ( a^2 = 1 ) implies ( a = 1 )2. ( 2ab = 1 ) implies ( b = frac{1}{2} )3. ( 2ac + b^2 - d^2 = 1 )4. ( 2bc - 2de = 1 )5. ( c^2 - e^2 = 1 )Solving these equations step-by-step:- From equation 1, ( a = 1 )- From equation 2, ( b = frac{1}{2} )- Substituting ( a = 1 ) and ( b = frac{1}{2} ) into equation 3: ( 2c + frac{1}{4} - d^2 = 1 ) leading to ( 2c - d^2 = frac{3}{4} )- From equation 4: ( c - 2de = 1 )- From equation 5: ( c^2 - e^2 = 1 )By solving these equations, we find:- ( e = 0 ), leading to ( c = 1 )- Substituting ( c = 1 ) into equation 3: ( 2(1) - d^2 = frac{3}{4} ) leading to ( d^2 = frac{5}{4} ) or ( d = frac{sqrt{5}}{2} )Thus, the polynomials are:[ A(x) = x^2 + frac{1}{2}x + 1 ][ B(x) = frac{sqrt{5}}{2}x ]Verifying the difference of squares:[ left( x^2 + frac{1}{2}x + 1 right)^2 - left( frac{sqrt{5}}{2}x right)^2 = x^4 + x^3 + x^2 + x + 1 ]The final answer is:[boxed{left(x^2 + frac{1}{2}x + 1right)^2 - left(frac{sqrt{5}}{2}xright)^2}],<think>Okay, so I need to express the polynomial ( x^4 + x^3 + x^2 + x + 1 ) as the difference of squares of two real-coefficient polynomials. And they have to be of different degrees. Hmm, interesting problem. Let me think about how to approach this.First, I know that a difference of squares is ( A^2 - B^2 ), which factors into ( (A - B)(A + B) ). But in this case, they want it expressed as the difference itself, not the factored form. So, I need to find two polynomials ( A ) and ( B ) such that ( A^2 - B^2 = x^4 + x^3 + x^2 + x + 1 ).Since the given polynomial is a quartic (degree 4), the highest degree term in the difference of squares must come from ( A^2 ) or ( B^2 ). Let's consider the degrees. If ( A ) is of degree 2, then ( A^2 ) would be degree 4. Similarly, if ( B ) is of degree less than 2, say 1, then ( B^2 ) would be degree 2. Then the difference ( A^2 - B^2 ) would have a leading term of degree 4. That seems plausible. But let's check if this works.Alternatively, if ( A ) is degree 3 and ( B ) is degree 2, then ( A^2 ) is degree 6 and ( B^2 ) is degree 4. But the given polynomial is degree 4, so the leading term must cancel out. That might not be helpful. So maybe the first idea is better: ( A ) is degree 2, ( B ) is degree 1 or 2. Wait, but they have to be different degrees. So if ( A ) is degree 2, then ( B ) has to be degree different, say 1 or 3. But if ( B ) is degree 3, then ( B^2 ) is degree 6, which when subtracted from ( A^2 ) (degree 4) would result in a negative degree 6 term, which isn't present in the original polynomial. So that's not going to work. So perhaps ( A ) is degree 2, ( B ) is degree 1. Let's try that.Let me denote ( A = ax^2 + bx + c ) and ( B = dx + e ). Then, ( A^2 = a^2x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2 ). ( B^2 = d^2x^2 + 2dex + e^2 ). Then, ( A^2 - B^2 = a^2x^4 + 2abx^3 + (2ac + b^2 - d^2)x^2 + (2bc - 2de)x + (c^2 - e^2) ).We need this to equal ( x^4 + x^3 + x^2 + x + 1 ). So, we can set up equations for the coefficients:1. Coefficient of ( x^4 ): ( a^2 = 1 )2. Coefficient of ( x^3 ): ( 2ab = 1 )3. Coefficient of ( x^2 ): ( 2ac + b^2 - d^2 = 1 )4. Coefficient of ( x ): ( 2bc - 2de = 1 )5. Constant term: ( c^2 - e^2 = 1 )Let me try solving these equations step by step.From equation 1: ( a^2 = 1 ) implies ( a = 1 ) or ( a = -1 ). Let's take ( a = 1 ) first. If that leads to a contradiction, we can try ( a = -1 ).Then equation 2: ( 2ab = 1 ). Since ( a = 1 ), this gives ( 2b = 1 implies b = 1/2 ).Now, equation 5: ( c^2 - e^2 = 1 ). Let's keep that in mind.Equation 4: ( 2bc - 2de = 1 ). We know ( b = 1/2 ), so this becomes ( 2*(1/2)*c - 2de = 1 implies c - 2de = 1 ).Equation 3: ( 2ac + b^2 - d^2 = 1 ). Again, ( a = 1 ), ( b = 1/2 ), so this becomes ( 2*1*c + (1/2)^2 - d^2 = 1 implies 2c + 1/4 - d^2 = 1 implies 2c - d^2 = 3/4 ).So, now we have three equations:From equation 5: ( c^2 - e^2 = 1 )From equation 4: ( c - 2de = 1 )From equation 3: ( 2c - d^2 = 3/4 )So we have three equations with variables c, d, e. Let me see if we can solve these.Let me denote equations:Equation 3: ( 2c - d^2 = 3/4 ). Let's solve for d^2: ( d^2 = 2c - 3/4 ). So d is real, so 2c - 3/4 must be non-negative. Therefore, ( 2c geq 3/4 implies c geq 3/8 ).Equation 4: ( c - 2de = 1 ). Let's solve for e: ( 2de = c - 1 implies e = (c - 1)/(2d) ). Assuming d ≠ 0.Equation 5: ( c^2 - e^2 = 1 ). Substitute e from equation 4 into this:( c^2 - [(c - 1)/(2d)]^2 = 1 )But from equation 3, ( d^2 = 2c - 3/4 implies d = sqrt(2c - 3/4) ). Let's take d positive for simplicity, since d is squared in equation 3. So, d = sqrt(2c - 3/4). Then, d^2 = 2c - 3/4.Therefore, substitute d into equation for e:( e = (c - 1)/(2*sqrt(2c - 3/4)) )Then, equation 5 becomes:( c^2 - [(c - 1)^2)/(4*(2c - 3/4))] = 1 )Let me write this out:( c^2 - frac{(c - 1)^2}{4*(2c - 3/4)} = 1 )Multiply both sides by 4*(2c - 3/4) to eliminate the denominator:4*(2c - 3/4)*c^2 - (c - 1)^2 = 4*(2c - 3/4)*1Let me compute each term:First term: 4*(2c - 3/4)*c^2 = 4c^2*(2c - 3/4) = 8c^3 - 3c^2Second term: - (c^2 - 2c + 1) = -c^2 + 2c -1Third term: 4*(2c - 3/4) = 8c - 3Putting it all together:8c^3 - 3c^2 - c^2 + 2c -1 = 8c - 3Simplify left side:8c^3 -4c^2 + 2c -1 = 8c -3Bring all terms to left side:8c^3 -4c^2 + 2c -1 -8c +3 =0Simplify:8c^3 -4c^2 -6c +2 =0So, the equation reduces to:8c^3 -4c^2 -6c +2 =0Hmm, solving this cubic equation. Let's see if we can factor this.Trying rational roots. The possible rational roots are ±1, ±2, ±1/2, ±1/4, ±1/8.Test c=1: 8 -4 -6 +2=0. 0! So c=1 is a root.Therefore, we can factor (c -1) from the cubic:Using polynomial division or synthetic division.Dividing 8c^3 -4c^2 -6c +2 by (c -1):Set up synthetic division:1 | 8 -4 -6 2 8 4 -2 8 4 -2 0So, the cubic factors as (c -1)(8c^2 +4c -2).Therefore, 8c^3 -4c^2 -6c +2= (c -1)(8c^2 +4c -2)Set equal to zero:(c -1)(8c^2 +4c -2)=0So solutions are c=1 and roots of 8c^2 +4c -2=0.Solve 8c^2 +4c -2=0 using quadratic formula:c = [-4 ± sqrt(16 + 64)] / 16 = [-4 ± sqrt(80)] /16 = [-4 ± 4*sqrt(5)] /16 = [-1 ± sqrt(5)]/4So, the roots are c=1, c=(-1 + sqrt(5))/4, and c=(-1 - sqrt(5))/4.Now, check which of these roots satisfy the earlier condition that c >= 3/8 (from equation 3: d^2 =2c -3/4 >=0).c=1: 2*1 -3/4= 5/4 >0, okay.c=(-1 + sqrt(5))/4: approximate value sqrt(5)≈2.236, so c≈(-1 +2.236)/4≈1.236/4≈0.309. 3/8=0.375. So 0.309 <0.375, so 2c -3/4=2*0.309 -0.75≈0.618 -0.75≈-0.132 <0. Not acceptable.Similarly, c=(-1 - sqrt(5))/4≈(-1 -2.236)/4≈-3.236/4≈-0.809, which is negative. 2c -3/4 would be even more negative. So discard these roots.Therefore, the only valid solution is c=1.So, c=1.Then from equation 3: d^2=2*1 -3/4=2 -0.75=1.25=5/4. So d= sqrt(5/4)= (√5)/2. Let's take d positive, so d=√5/2.Then from equation 4: e=(c -1)/(2d)= (1 -1)/(2*(√5/2))=0/(√5)=0. So e=0.From equation 5: c^2 - e^2=1^2 -0=1, which checks out.So, putting it all together:a=1, b=1/2, c=1, d=√5/2, e=0.Therefore, the polynomials are:A= ax^2 +bx +c= x^2 + (1/2)x +1B= dx +e= (√5/2)x +0= (√5/2)xSo, the difference of squares is:A^2 - B^2 = (x^2 + (1/2)x +1)^2 - ( (√5/2 x)^2 )Let me compute this to verify.First, compute A^2:(x^2 + (1/2)x +1)^2 = x^4 + (1/2)x^3 +1x^2 + (1/2)x^3 + (1/4)x^2 + (1/2)x +1x^2 + (1/2)x +1Wait, perhaps better to do step by step:= (x^2)^2 + 2*x^2*(1/2 x) + 2*x^2*1 + (1/2 x)^2 + 2*(1/2 x)*1 +1^2Wait, no. The standard expansion (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc.So here, a=x^2, b=(1/2)x, c=1.Thus:= (x^2)^2 + ((1/2)x)^2 +1^2 + 2*x^2*(1/2x) + 2*x^2*1 + 2*(1/2x)*1Compute term by term:(x^2)^2 =x^4((1/2)x)^2= (1/4)x^21^2=12*x^2*(1/2x)= 2*(1/2)x^3= x^32*x^2*1=2x^22*(1/2x)*1= (2*(1/2))x= xSo adding all together:x^4 + (1/4)x^2 +1 +x^3 +2x^2 +xCombine like terms:x^4 +x^3 + (1/4x^2 +2x^2) +x +1Which is:x^4 +x^3 + (9/4)x^2 +x +1Now compute B^2 = ( (√5/2 x )^2 ) = (5/4)x^2So, A^2 - B^2 = [x^4 +x^3 + (9/4)x^2 +x +1] - (5/4)x^2 =x^4 +x^3 + (9/4 -5/4)x^2 +x +1= x^4 +x^3 +x^2 +x +1. Which is exactly the polynomial we needed! So that works.Therefore, the answer is ( left( x^2 + frac{1}{2}x + 1 right)^2 - left( frac{sqrt{5}}{2}x right)^2 ).But the problem says "as the difference of squares of two real-coefficient polynomials of different degrees." So, A is degree 2 and B is degree 1, which are different. So that's good.Alternatively, we can write it as:( left( x^2 + frac{1}{2}x + 1 right)^2 - left( frac{sqrt{5}}{2}x right)^2 )But maybe they want integer coefficients? Wait, the problem says real coefficients, so fractions and sqrt(5) are okay.Alternatively, we can rationalize the coefficients by scaling A and B, but I think the above is acceptable.Alternatively, factor out the 1/2 coefficients. Wait, but in the difference of squares, scaling would affect the squares. Let me check if there's a way to write it with integer coefficients, but given the sqrt(5), probably not. Since sqrt(5) is irrational, we can't have integer coefficients. So, the answer must involve fractions and sqrt(5).Alternatively, maybe another approach could yield different polynomials without sqrt(5), but I need to check.Wait, another thought. The given polynomial is the 5th cyclotomic polynomial: ( x^4 +x^3 +x^2 +x +1 = frac{x^5 -1}{x -1} ). Not sure if that helps here.Alternatively, maybe factor the polynomial into quadratics and see if it can be written as difference of squares. Let's see.But the polynomial ( x^4 +x^3 +x^2 +x +1 ) does not factor into quadratics with real coefficients. Wait, does it?Let me try to factor it. Suppose it factors as (x^2 + ax + b)(x^2 + cx + d). Then expanding:x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd.Set equal to x^4 +x^3 +x^2 +x +1. Therefore:1. a + c =12. ac + b + d =13. ad + bc =14. bd =1We need to solve this system. Let me try.From equation 4: bd=1. So possible real solutions: b=1, d=1 or b=-1, d=-1.First, try b=1, d=1.Then, equation 1: a + c =1Equation 2: ac +1 +1=1 => ac= -1Equation 3: a*1 + c*1= a + c=1. Wait, equation 3 is a + c=1. But equation 1 is a + c=1, so equation 3 is redundant. So, we have:From equation 1 and 2: a + c=1 and ac=-1.So, solving for a and c: They are roots of t^2 - t -1=0. Which gives t=(1 ± sqrt(5))/2. So a=(1 + sqrt(5))/2, c=(1 - sqrt(5))/2 or vice versa.Therefore, the polynomial factors into (x^2 + [(1 + sqrt(5))/2]x +1)(x^2 + [(1 - sqrt(5))/2]x +1). So it factors into two quadratics with real coefficients.But how does that help us with expressing it as a difference of squares? Hmm.If the polynomial can be factored into two quadratics, maybe we can use the identity ( (A + B)(A - B) = A^2 - B^2 ). But here, the factors are not a sum and difference, but two different quadratics. So perhaps if we can set one quadratic as A + B and the other as A - B, but they are not symmetric.Alternatively, if we can write the polynomial as a product of two quadratics and then express each quadratic as a square or something else. But I'm not sure. Maybe this approach is more complicated.Alternatively, going back to the first approach where we found A and B, that seems to work. So maybe that's the way to go.So, the answer is ( (x^2 + frac{1}{2}x +1)^2 - (frac{sqrt{5}}{2}x)^2 ). Let me check once more:Compute ( (x^2 + frac{1}{2}x +1)^2 ):First square: x^4 + x^3 + (1/4 + 2)x^2 + x +1, which when expanded as before gives x^4 +x^3 + (9/4)x^2 +x +1.Subtract ( (frac{sqrt{5}}{2}x)^2 = frac{5}{4}x^2 ), so 9/4x^2 -5/4x^2= x^2. Therefore, result is x^4 +x^3 +x^2 +x +1. Correct.Therefore, this works. The polynomials A and B are of degrees 2 and 1 respectively, which are different.Thus, the required expression is ( left( x^2 + frac{1}{2}x + 1 right)^2 - left( frac{sqrt{5}}{2}x right)^2 ).Final AnswerThe polynomial can be expressed as the difference of squares: boxed{left(x^2 + frac{1}{2}x + 1right)^2 - left(frac{sqrt{5}}{2}xright)^2}.</think>To express the polynomial ( x^4 + x^3 + x^2 + x + 1 ) as the difference of squares of two real-coefficient polynomials of different degrees, we start by considering polynomials ( A ) and ( B ) such that ( A^2 - B^2 = x^4 + x^3 + x^2 + x + 1 ).We assume ( A ) is a quadratic polynomial and ( B ) is a linear polynomial. Let ( A = ax^2 + bx + c ) and ( B = dx + e ). Expanding ( A^2 ) and ( B^2 ), we equate the coefficients of the resulting polynomial to those of ( x^4 + x^3 + x^2 + x + 1 ).1. Coefficient of ( x^4 ): ( a^2 = 1 ) implies ( a = 1 ).2. Coefficient of ( x^3 ): ( 2ab = 1 ) implies ( b = frac{1}{2} ).3. Coefficient of ( x^2 ): ( 2ac + b^2 - d^2 = 1 ).4. Coefficient of ( x ): ( 2bc - 2de = 1 ).5. Constant term: ( c^2 - e^2 = 1 ).Solving these equations step-by-step, we find:- From ( a = 1 ) and ( b = frac{1}{2} ), we substitute into the other equations.- Solving for ( c ), ( d ), and ( e ), we find ( c = 1 ), ( d = frac{sqrt{5}}{2} ), and ( e = 0 ).Thus, the polynomials are ( A = x^2 + frac{1}{2}x + 1 ) and ( B = frac{sqrt{5}}{2}x ). Verifying the difference of squares:- ( A^2 = (x^2 + frac{1}{2}x + 1)^2 = x^4 + x^3 + frac{9}{4}x^2 + x + 1 )- ( B^2 = (frac{sqrt{5}}{2}x)^2 = frac{5}{4}x^2 )- Subtracting these, we get ( x^4 + x^3 + x^2 + x + 1 )Therefore, the polynomial ( x^4 + x^3 + x^2 + x + 1 ) can be expressed as the difference of squares:[boxed{left(x^2 + frac{1}{2}x + 1right)^2 - left(frac{sqrt{5}}{2}xright)^2}]]

🔑:[Solution] If the polynomial W(x)=x^{4}+x^{3}+x^{2}+x+1 is expressed as U^{2}(x) - V^{2}(x), where U(x) and V(x) are two polynomials of different degrees, then the polynomial U(x) must be quadratic, and the polynomial V(x) must be linear or constant. Therefore, the polynomial V^{2}(x) does not contain terms of degree higher than 2, which means that the highest degree and the next highest degree terms of the polynomials U^{2}(x) and W(x) must be the same, so U^{2}(x) has the form x^{4}+x^{3}+cdots. Thus, U(x) can be expressed as U(x)=x^{2}+frac{1}{2} x+a. Therefore,begin{aligned}V^{2}(x)=U^{2}(x)-W(x)= & left(x^{2}+frac{1}{2} x+aright)^{2}-left(x^{4}+x^{3}+x^{2}+xright. & +1) = & left(2 a-frac{3}{4}right) x^{2}+(a-1) x+a^{2}-1 .end{aligned}Since 2 a-frac{3}{4}=0 and a-1=0 cannot hold simultaneously, it follows from the expression of V^{2}(x) that V(x) cannot be a constant. Therefore, V(x) must be a linear polynomial. This indicates that V^{2}(x) is a quadratic trinomial with a positive leading coefficient and a discriminant of zero, i.e.,2 a-frac{3}{4}>0, (a-1)^{2}-4left(2 a-frac{3}{4}right)left(a^{2}-1right)=0 text { . }The second condition can be simplified to(a-1)left(4 a^{2}+2 a-1right)=0 text {, }Solving this gives a=1, a=frac{1}{4}(-1 pm sqrt{5}). Among these, only a=1 satisfies the condition2 a-frac{3}{4}>0 text { . }Thus, the original polynomial can be expressed asx^{4}+x^{3}+x^{2}+x+1=left(x^{2}+frac{1}{2} x+1right)^{2}-left(frac{1}{2} sqrt{5} xright)^{2} text {, }and the representation is unique.