🧠:[<think>Okay, so I need to figure out this problem where I have to multiply 4.02 by a two-digit integer, and the product has to be an integer. Then, 10 times that product is the answer they want. Let me break this down step by step.First, the problem says to multiply 4.02 by a two-digit integer. Let me denote that two-digit integer as 'n'. So, the equation would be 4.02 * n = integer. Then, 10 times that integer is the final answer we need to find. Hmm. Wait, so actually, if I first find such an 'n' that when multiplied by 4.02 gives an integer, then multiply that integer by 10, and that's the answer?Let me confirm. The problem says, "Multiply 4.02 by a two-digit integer, and the product is an integer. Then, 10 times this product is ______." So yes, the blank is filled with 10*(4.02*n), where n is a two-digit integer making 4.02*n an integer. So maybe the question is asking for the value of 10*(4.02*n), given that n is a two-digit integer and 4.02*n is integer.Alternatively, maybe the problem is phrased such that 10 times the product (which is 10*(4.02*n)) is the answer they want. So regardless, I need to find such an n and compute 10*(4.02*n). But perhaps there might be multiple possible n's? But the problem doesn't specify to find all possible answers, so maybe there's a unique answer? Let me check.Wait, 4.02 is a decimal number. If we can write it as a fraction, that might help. 4.02 is equal to 402/100, which simplifies to 201/50. Because 402 divided by 2 is 201, and 100 divided by 2 is 50. So 4.02 = 201/50. Therefore, multiplying 201/50 by n should result in an integer. So, (201/50)*n must be an integer. Therefore, 201n must be divisible by 50. So, 201n ≡ 0 mod 50. Therefore, 201n is a multiple of 50.Since 201 and 50 are co-prime or not? Let's check the greatest common divisor (GCD) of 201 and 50. Prime factors of 201: 3*67. Prime factors of 50: 2*5^2. No common factors, so GCD(201,50)=1. Therefore, in order for 201n to be divisible by 50, n must be divisible by 50. Because since 201 and 50 are coprime, the n has to supply the factors needed for 50. So, n must be a multiple of 50.But n is a two-digit integer. The two-digit integers are from 10 to 99 inclusive. So the multiples of 50 in that range are 50, 100, etc., but 100 is a three-digit number, so only 50 is the multiple of 50 that is a two-digit integer. Therefore, n must be 50.Therefore, n=50. Then, 4.02*50 = ?Calculating 4.02*50: 4*50=200, 0.02*50=1, so total is 201. Therefore, the product is 201. Then, 10 times that product is 201*10=2010. Therefore, the answer is 2010.Wait, but let me check if there are other possible two-digit integers n that could make 4.02*n an integer, even if n is not a multiple of 50. Since maybe 201n divisible by 50 doesn't necessarily require n to be a multiple of 50, but perhaps n has some factors that when multiplied by 201, the product has factors of 2 and 5^2. Let me think again.201 is 3*67, and 50 is 2*5^2. Since 201 and 50 are coprime, as established before, the only way 201n is divisible by 50 is if n is divisible by 50. Because 201 doesn't contribute any factors of 2 or 5, so n must contribute all the factors of 2 and 5 required for 50.Therefore, n must be divisible by 50, and since n is a two-digit number, n=50 is the only possibility. So the product is 4.02*50=201, and 10 times that is 2010. Therefore, the answer is 2010. Let me check again.Alternative approach: 4.02 is two decimal places, so when multiplied by a number, the product will be an integer if the two decimal places are canceled out. So, multiplying by 100 would shift the decimal two places, making it an integer. But here, we need to multiply by a two-digit integer. So, 4.02 * n = integer. The decimal part is 0.02, so 0.02 * n must be an integer. Therefore, 0.02n must be integer. 0.02 is 1/50, so 1/50 * n is integer. Therefore, n must be a multiple of 50. So again, n=50 is the only two-digit multiple of 50. Therefore, same conclusion.Therefore, the product is 4.02*50=201, and 10 times that is 2010. Hence, the answer is 2010. Seems solid.Final Answerboxed{2010}</think>To solve the problem, we need to multiply 4.02 by a two-digit integer such that the product is an integer. Then, we need to find 10 times this product.First, we convert 4.02 into a fraction:[4.02 = frac{402}{100} = frac{201}{50}]We need to find a two-digit integer ( n ) such that:[frac{201}{50} times n text{ is an integer}]This implies that ( 201n ) must be divisible by 50. Since 201 and 50 are coprime (their greatest common divisor is 1), ( n ) must be a multiple of 50. The only two-digit multiple of 50 is 50 itself.Thus, we use ( n = 50 ):[4.02 times 50 = 201]Next, we calculate 10 times this product:[10 times 201 = 2010]Therefore, the final answer is:[boxed{2010}],<think>Okay, let me try to figure out this problem. So, the question is: Multiply 4.02 by a two-digit integer, and the product is an integer. Then, 10 times this product is... something. Hmm, the question is asking for what 10 times this product is, but I think the key part is first figuring out what that two-digit integer is. Let me break it down step by step.First, we have 4.02 multiplied by a two-digit integer, let's call that integer "n". The result of this multiplication is an integer. Then, we need to take that product and multiply it by 10, and the question is asking for the value of that final number. So, essentially, they want 10*(4.02*n), where n is a two-digit integer such that 4.02*n is an integer. But maybe the question is phrased differently. Let me check again: "Multiply 4.02 by a two-digit integer, and the product is an integer. Then, 10 times this product is qquad ." So, perhaps they just want 10 times the product, given that the product is an integer. But how do we find what that product is? There must be a specific answer here. Wait, maybe there are multiple possible answers, but since the problem is worth 8 points, maybe there's a unique answer. Hmm, not sure. Let me think.First, 4.02 is a decimal number. If we multiply it by a two-digit integer n to get an integer product, that means 4.02 must be a divisor of some integer, and n is such that when multiplied by 4.02, the result is integer. Alternatively, 4.02 can be expressed as a fraction. Let me convert 4.02 into a fraction. Since 4.02 is 4 and 2 hundredths, that's 4 + 2/100 = 4 + 1/50 = 201/50. So 4.02 is equal to 201/50. So, if we write 4.02 * n = integer, then substituting the fraction, we have (201/50)*n must be an integer. So, (201n)/50 must be integer, which implies that 50 divides 201n. Since 201 and 50 are co-prime? Let me check. The prime factors of 50 are 2*5*5. The prime factors of 201 are 3*67. So yes, 201 and 50 are co-prime because they share no common prime factors. Therefore, for 50 to divide 201n, 50 must divide n. Because 201 and 50 are co-prime, so n must be a multiple of 50. But n is a two-digit integer. The two-digit integers are from 10 to 99. But 50 is a two-digit integer. The multiples of 50 in the two-digit range are 50, 100, but 100 is three digits, so only 50. Therefore, n must be 50. Wait, that seems too straightforward. Let me check. If n is 50, then 4.02 * 50 = 4.02 * 50. Let's compute that. 4 * 50 = 200, and 0.02 * 50 = 1, so total is 201. So the product is 201, which is an integer. Then 10 times this product is 2010. So is the answer 2010? But the problem says "a two-digit integer". If n must be a two-digit integer, then 50 is the only possible value? Let me confirm.Alternatively, maybe there's another approach. Let's think: 4.02 has two decimal places. Multiplying by a two-digit integer n, so that the product is an integer. So the two decimal places must be canceled out by the multiplication. So, 4.02 is equivalent to 402/100. Therefore, multiplying by n gives (402/100)*n. For this to be integer, 100 must divide 402n. So, 402n must be divisible by 100. Simplifying, 402 and 100 have a greatest common divisor (GCD). Let me compute GCD(402,100). The prime factors of 402: 2*3*67. The prime factors of 100: 2^2*5^2. So the GCD is 2. Therefore, 402n/100 reduces to (201n)/50. For 201n/50 to be integer, 50 must divide 201n. Since GCD(201,50)=1, as before, 50 must divide n. Therefore, n must be a multiple of 50. The only two-digit multiple is 50. Therefore, n=50, so the product is 201, and 10 times that is 2010. So the answer is 2010. But let me verify once more. Suppose someone thinks, "Wait, maybe there's another two-digit number that works?" Let's test with n=25. 4.02*25. 4*25=100, 0.02*25=0.5, total is 100.5, which is not integer. How about n=20? 4.02*20=80.4, not integer. n=10: 40.2. n=50: 201. n=75: 4.02*75=301.5, which is not integer. So indeed, only n=50 gives an integer product. Therefore, 10 times 201 is 2010. So the answer should be 2010. Alternatively, perhaps the problem is designed such that even if there are multiple possible n's, but 10 times the product is always the same? Let's check. Suppose there's another n where 4.02*n is integer, but maybe it's not a two-digit number. Wait, but the problem specifies that n is a two-digit integer, so the only possible n is 50. Therefore, the answer must be 2010. Wait a second, but maybe the problem is expecting 10 times the product, but they just need to know that 4.02* n *10 = 40.2*n, which would be an integer since 4.02*n is an integer. But the problem says "Multiply 4.02 by a two-digit integer, and the product is an integer. Then, 10 times this product is ___." So regardless of what the product is, 10 times it would be an integer, but the problem is perhaps expecting a numerical answer. But since n is fixed as 50, the answer is 2010. But is 2010 the only possible answer?Wait, suppose n is a two-digit number such that 4.02*n is integer. Then 10*(4.02*n) is 40.2*n. But 40.2 is 402/10, so 402/10 *n. But since 4.02*n is integer, which is 402/100 *n integer. Then 402n must be divisible by 100. Which again requires n divisible by 50. So n=50. Therefore, 10 times that is 402*50/10=2010. Alternatively, let me do this with equations. Let me write:Let n be a two-digit integer such that 4.02 * n is integer.4.02 = 402/100, so 402/100 * n = integer.Therefore, 402n must be divisible by 100. 402 and 100 have GCD 2, so 402n divisible by 100 implies that (402/2)n = 201n must be divisible by 50. Since GCD(201,50)=1, 201 and 50 are coprime, so 50 divides n. Therefore, n must be divisible by 50. The only two-digit number divisible by 50 is 50. Therefore, n=50.Then, 4.02 *50=201, and 10*201=2010. Therefore, the answer is 2010.But let me check if there's another way. For example, suppose n is 25, which is two-digit. Then 4.02*25 is 100.5, which is not integer. If n=40, 4.02*40=160.8, not integer. n=30: 120.6, not integer. n=50: 201, which is integer. So yes, only n=50. Therefore, 10*201=2010. Hence, the answer is 2010. I think that's solid. The key steps are converting 4.02 into a fraction, recognizing that the denominator must divide the product of the numerator and the integer n, and since the denominator is 100 (from 402/100), and simplifying the divisibility condition leads us to n must be a multiple of 50, but since n is two-digit, only 50 works. Therefore, the product is 201, and 10 times that is 2010. Final AnswerThe final answer is boxed{2010}.</think>To solve the problem, we need to multiply 4.02 by a two-digit integer such that the product is an integer, and then find 10 times this product.1. Convert 4.02 to a fraction: [ 4.02 = frac{402}{100} = frac{201}{50} ]2. Set up the multiplication with a two-digit integer ( n ): [ frac{201}{50} times n text{ must be an integer} ]3. Determine the condition for ( n ): - For (frac{201n}{50}) to be an integer, 50 must divide (201n). - Since 201 and 50 are coprime (their greatest common divisor is 1), ( n ) must be a multiple of 50.4. Identify the two-digit integer ( n ): - The only two-digit multiple of 50 is 50 itself.5. Calculate the product: [ 4.02 times 50 = 201 ]6. Find 10 times the product: [ 10 times 201 = 2010 ]Thus, the final answer is (boxed{2010}).]

🔑:【Answer】Solution: 4.02 multiplied by a two-digit integer, the product is an integer, this two-digit number is 50, 4.02 times 50 times 10=2010.Answer: The 10 times of this product is 2010.Therefore, the answer is: 2010.

🧠:[<think>Okay, so I need to solve this equation: √(3x² - 2x + 15) + √(3x² - 2x + 8) = 7. Hmm, let's see. First, I remember that when dealing with square roots, sometimes squaring both sides can help eliminate the radicals. But there are two square roots here, so squaring might get a bit messy. Let me think step by step.Let me denote the two square roots as separate terms to simplify notation. Let’s set A = √(3x² - 2x + 15) and B = √(3x² - 2x + 8). Then the equation becomes A + B = 7. If I square both sides, I can use the formula (A + B)² = A² + 2AB + B². Let me try that:(A + B)² = 7² A² + 2AB + B² = 49But A² is (3x² - 2x + 15) and B² is (3x² - 2x + 8). Let's substitute those in:(3x² - 2x + 15) + 2AB + (3x² - 2x + 8) = 49Combine like terms:3x² + 3x² = 6x² -2x -2x = -4x 15 + 8 = 23So, 6x² - 4x + 23 + 2AB = 49Now, subtract 23 from both sides:6x² - 4x + 2AB = 26Hmm, but there's still that 2AB term. AB is √(3x² - 2x + 15) * √(3x² - 2x + 8). That looks complicated. Maybe I can express AB in terms of the original expressions?Alternatively, maybe I can isolate one of the square roots first before squaring. Let me try that approach. Let's suppose I move one of the radicals to the other side:√(3x² - 2x + 15) = 7 - √(3x² - 2x + 8)Now, square both sides:(√(3x² - 2x + 15))² = (7 - √(3x² - 2x + 8))² 3x² - 2x + 15 = 49 - 14√(3x² - 2x + 8) + (3x² - 2x + 8)Simplify the right-hand side:49 + (3x² - 2x + 8) = 3x² - 2x + 57 So, 3x² - 2x + 15 = 3x² - 2x + 57 - 14√(3x² - 2x + 8)Subtract 3x² - 2x from both sides:15 = 57 - 14√(3x² - 2x + 8)Now, subtract 57 from both sides:15 - 57 = -14√(3x² - 2x + 8) -42 = -14√(3x² - 2x + 8)Divide both sides by -14:3 = √(3x² - 2x + 8)Now, square both sides again:9 = 3x² - 2x + 8Subtract 9 from both sides:0 = 3x² - 2x -1So, 3x² - 2x -1 = 0Now, solving this quadratic equation. Let's use the quadratic formula. The quadratic is ax² + bx + c = 0, so a = 3, b = -2, c = -1.x = [-b ± √(b² - 4ac)] / (2a) x = [2 ± √(4 - 4*3*(-1))]/6 x = [2 ± √(4 + 12)]/6 x = [2 ± √16]/6 x = [2 ± 4]/6So, two solutions:x = (2 + 4)/6 = 6/6 = 1 x = (2 - 4)/6 = (-2)/6 = -1/3Now, we need to check these solutions in the original equation because squaring can sometimes introduce extraneous solutions.First, check x = 1:Compute each radical:First radical: √(3(1)² - 2(1) + 15) = √(3 - 2 + 15) = √16 = 4 Second radical: √(3(1)² - 2(1) + 8) = √(3 - 2 + 8) = √9 = 3 Sum: 4 + 3 = 7. Which matches the right-hand side. So, x = 1 is valid.Now, check x = -1/3:First radical: √(3*(-1/3)² - 2*(-1/3) + 15) Calculate term by term:3*(1/9) = 1/3 -2*(-1/3) = 2/3 15 remains Total inside radical: 1/3 + 2/3 + 15 = (1/3 + 2/3) + 15 = 1 + 15 = 16 So, first radical is √16 = 4Second radical: √(3*(-1/3)² - 2*(-1/3) + 8) Again, term by term:3*(1/9) = 1/3 -2*(-1/3) = 2/3 8 remains Total inside radical: 1/3 + 2/3 + 8 = (1/3 + 2/3) + 8 = 1 + 8 = 9 Second radical is √9 = 3Sum: 4 + 3 = 7. Which also matches. So, x = -1/3 is also valid.Wait, both solutions work? That's interesting. Let me double-check the computations for x = -1/3 because sometimes even if you do everything right, it's good to verify.First radical:3x² = 3*(1/9) = 1/3 -2x = -2*(-1/3) = 2/3 So 1/3 + 2/3 +15 = 3/3 +15 = 1 +15=16. Yep.Second radical:3x² = 1/3 -2x = 2/3 So 1/3 + 2/3 +8=1 +8=9. Correct.So both solutions are valid. Therefore, the solutions are x = 1 and x = -1/3.But wait, just to make sure, let's check if there were any steps where we might have introduced extraneous solutions. When we squared both sides, sometimes solutions can appear that don't satisfy the original equation because squaring is not a reversible operation. However, in this case, both solutions checked out when substituted back into the original equation. So, they are both valid.Therefore, the solutions are x = 1 and x = -1/3.Final AnswerThe solutions are boxed{1} and boxed{-dfrac{1}{3}}.</think>Given the equation (sqrt{3x^2 - 2x + 15} + sqrt{3x^2 - 2x + 8} = 7), we start by isolating one of the square roots:[sqrt{3x^2 - 2x + 15} = 7 - sqrt{3x^2 - 2x + 8}]Squaring both sides, we get:[3x^2 - 2x + 15 = 49 - 14sqrt{3x^2 - 2x + 8} + 3x^2 - 2x + 8]Simplifying the right-hand side:[3x^2 - 2x + 15 = 57 - 14sqrt{3x^2 - 2x + 8}]Subtracting (3x^2 - 2x) from both sides:[15 = 57 - 14sqrt{3x^2 - 2x + 8}]Subtracting 57 from both sides:[-42 = -14sqrt{3x^2 - 2x + 8}]Dividing by -14:[3 = sqrt{3x^2 - 2x + 8}]Squaring both sides again:[9 = 3x^2 - 2x + 8]Subtracting 9 from both sides:[0 = 3x^2 - 2x - 1]Solving the quadratic equation (3x^2 - 2x - 1 = 0) using the quadratic formula (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}) where (a = 3), (b = -2), and (c = -1):[x = frac{2 pm sqrt{4 + 12}}{6} = frac{2 pm 4}{6}]This gives the solutions:[x = frac{6}{6} = 1 quad text{and} quad x = frac{-2}{6} = -frac{1}{3}]Checking both solutions in the original equation confirms they are valid. Therefore, the solutions are:[boxed{1} quad text{and} quad boxed{-dfrac{1}{3}}],<think>Okay, let's see. I need to solve the equation: √(3x² - 2x + 15) + √(3x² - 2x + 8) = 7. Hmm, radicals can be tricky. Maybe I can simplify this by substitution. Let me check the expressions under the square roots. Both have 3x² - 2x, but with different constants: 15 and 8. Maybe if I set a variable to represent that common quadratic part. Let's try letting y = 3x² - 2x. Then the equation becomes √(y + 15) + √(y + 8) = 7. That looks simpler. So now I have √(y + 15) + √(y + 8) = 7. Maybe I can let one of the square roots be a variable. Let me set, say, a = √(y + 15), then the other term would be √(y + 8) = 7 - a. Then I can square both sides to eliminate the square roots. Let's try that.First, set a = √(y + 15). Then √(y + 8) = 7 - a. Now square both sides: (√(y + 8))² = (7 - a)². That gives y + 8 = 49 - 14a + a². But since a = √(y + 15), then a² = y + 15. Substitute that into the equation: y + 8 = 49 - 14a + (y + 15). Let's simplify the right side: 49 + 15 = 64, so it's y + 64 - 14a. Now the equation is y + 8 = y + 64 - 14a. Subtract y from both sides: 8 = 64 - 14a. Then subtract 64 from both sides: 8 - 64 = -14a => -56 = -14a. Divide both sides by -14: a = 4. So a = 4. Recall that a = √(y + 15), so 4 = √(y + 15). Square both sides: 16 = y + 15. Therefore, y = 1. Now, since y = 3x² - 2x, substitute back: 3x² - 2x = 1. Then 3x² - 2x - 1 = 0. Let's solve this quadratic equation. Using the quadratic formula: x = [2 ± √(4 + 12)] / 6. Wait, discriminant is 4 + 12 = 16. So √16 = 4. Therefore, x = [2 ± 4]/6. That gives two solutions: (2 + 4)/6 = 6/6 = 1, and (2 - 4)/6 = (-2)/6 = -1/3. But wait, we need to check these solutions in the original equation because sometimes squaring can introduce extraneous solutions. Let me check x = 1 first. Substitute x = 1 into the original equation:First radical: √(3*(1)^2 - 2*(1) + 15) = √(3 - 2 + 15) = √16 = 4.Second radical: √(3*(1)^2 - 2*(1) + 8) = √(3 - 2 + 8) = √9 = 3.So 4 + 3 = 7, which matches the right side. So x = 1 is valid.Now check x = -1/3. Let's compute each radical:First radical: 3*(-1/3)^2 - 2*(-1/3) + 15. Let's compute term by term.3*(1/9) = 1/3-2*(-1/3) = 2/3So adding them up: 1/3 + 2/3 + 15 = (1/3 + 2/3) + 15 = 1 + 15 = 16. So the first radical is √16 = 4.Second radical: 3*(-1/3)^2 - 2*(-1/3) + 8.Again, 3*(1/9) = 1/3-2*(-1/3) = 2/3So total: 1/3 + 2/3 + 8 = 1 + 8 = 9. So √9 = 3.Thus, 4 + 3 = 7, which also holds. So x = -1/3 is also a valid solution.Therefore, both solutions are valid. But wait, the original equation had square roots, which require the expressions under them to be non-negative. Let me confirm that for both x = 1 and x = -1/3, the expressions under the radicals are non-negative.For x = 1:3x² - 2x + 15 = 3 - 2 + 15 = 16 ≥ 0.3x² - 2x + 8 = same but 8 instead of 15: 3 - 2 + 8 = 9 ≥ 0.For x = -1/3:3x² = 3*(1/9) = 1/3-2x = -2*(-1/3) = 2/3So 3x² - 2x = 1/3 + 2/3 = 1.Thus, 1 + 15 = 16 and 1 + 8 = 9, both positive. So both expressions under radicals are positive, so both solutions are valid.Therefore, the solutions are x = 1 and x = -1/3.But let me check if there's another way to approach the problem, maybe directly without substitution. Let's try that.Original equation: √(3x² - 2x + 15) + √(3x² - 2x + 8) = 7.Let’s denote the two radicals as A and B. So A + B = 7. If we let A = √(3x² - 2x +15) and B = √(3x² - 2x +8). Then A - B = (A + B)(A - B) / (A + B) = (A² - B²)/ (A + B). Compute A² - B²: (3x² -2x +15) - (3x² -2x +8) = 7. So A² - B² = 7. Then (A - B) = 7 / (A + B) = 7 / 7 = 1. Therefore, A - B = 1. So now we have the system:A + B = 7A - B = 1Adding both equations: 2A = 8 => A = 4. Then B = 7 - A = 3. Which is the same result as before. So then A = 4, which gives 3x² -2x +15 = 16 => 3x² -2x -1 = 0. Same quadratic equation. So same solutions.So regardless of the method, we get the same quadratic equation. So the solutions are x = 1 and x = -1/3. Both checked and valid.Therefore, the answer is x = 1 and x = -1/3.Final AnswerThe solutions are boxed{1} and boxed{-dfrac{1}{3}}.</think>Given the equation (sqrt{3x^2 - 2x + 15} + sqrt{3x^2 - 2x + 8} = 7), we start by substituting (y = 3x^2 - 2x). This transforms the equation into (sqrt{y + 15} + sqrt{y + 8} = 7).Let (a = sqrt{y + 15}). Then the equation becomes (a + sqrt{y + 8} = 7). Solving for (sqrt{y + 8}), we get (sqrt{y + 8} = 7 - a). Squaring both sides, we have:[y + 8 = (7 - a)^2]Since (a = sqrt{y + 15}), squaring this gives (a^2 = y + 15). Substituting (a^2) into the equation:[y + 8 = 49 - 14a + y + 15]Simplifying, we get:[8 = 64 - 14a]Solving for (a):[-56 = -14a implies a = 4]Substituting back, (a = sqrt{y + 15} = 4) gives:[16 = y + 15 implies y = 1]Since (y = 3x^2 - 2x), we solve the quadratic equation (3x^2 - 2x - 1 = 0) using the quadratic formula:[x = frac{2 pm sqrt{4 + 12}}{6} = frac{2 pm 4}{6}]This gives the solutions:[x = frac{6}{6} = 1 quad text{and} quad x = frac{-2}{6} = -frac{1}{3}]Checking both solutions in the original equation:- For (x = 1), both radicals evaluate to 4 and 3, respectively, summing to 7.- For (x = -frac{1}{3}), both radicals evaluate to 4 and 3, respectively, summing to 7.Both solutions are valid. Thus, the solutions are (boxed{1}) and (boxed{-dfrac{1}{3}}).]

🔑:## Solution.Let 3 x^{2}-2 x=y. With respect to y, the equation takes the form sqrt{y+15}+sqrt{y+8}=7. Squaring both sides of the equation, we get y+15+2 sqrt{(y+15)(y+8)}+y+8=49 Leftrightarrow sqrt{(y+15)(y+8)}=13-y, where 13-y geq 0, y leq 13. From this, (y+15)(y+8)=(13-y)^{2}, y=1.Then 3 x^{2}-2 x=1,3 x^{2}-2 x-1=0, from which x_{1}=-frac{1}{3}, x_{2}=1.Answer: x_{1}=-frac{1}{3}, x_{2}=1.

🧠:[<think>Okay, so I need to figure out which of the given functions is both a decreasing function and an odd function. Let me recall what an odd function is. An odd function satisfies the property f(-x) = -f(x) for all x in its domain. And a decreasing function is one where, if x1 < x2, then f(x1) > f(x2). Alternatively, the derivative f’(x) is negative everywhere in its domain. Hmm, maybe checking the derivative could be a good approach here.Let me look at each option one by one.Starting with option A: f(x) = log₂(√(x² +1) - x). Let's see. First, is this an odd function? Let's compute f(-x). If I substitute -x into the function, I get log₂(√((-x)² +1) - (-x)) = log₂(√(x² +1) + x). Now, let's see if this is equal to -f(x). The original f(x) is log₂(√(x² +1) - x). Let's compute -f(x) = -log₂(√(x² +1) - x). Is log₂(√(x² +1) + x) equal to -log₂(√(x² +1) - x)?Hmm, maybe using logarithmic properties here. Let's note that (√(x² +1) - x)(√(x² +1) + x) = (√(x² +1))² - x² = x² +1 - x² = 1. So the product of (√(x² +1) - x) and (√(x² +1) + x) is 1. Therefore, log₂(√(x² +1) + x) = log₂(1/(√(x² +1) - x)) = -log₂(√(x² +1) - x). So that means f(-x) = -f(x), which confirms that it is indeed an odd function. So option A is odd. Now, is it decreasing?To check if it's decreasing, maybe take the derivative. Let me compute f'(x). Let f(x) = log₂(u), where u = √(x² +1) - x. The derivative of log₂(u) is 1/(u ln 2) * du/dx. Let's compute du/dx. u = √(x² +1) - x. The derivative of √(x² +1) is (1/(2√(x² +1)))*2x = x/√(x² +1). Then derivative of -x is -1. So du/dx = x/√(x² +1) - 1. Therefore, f'(x) = [x/√(x² +1) - 1] / [ ( √(x² +1) - x ) ln 2 ].Simplify numerator: x/√(x² +1) - 1. Let's write it as (x - √(x² +1))/√(x² +1). Because if I factor out 1/√(x² +1), then x - √(x² +1). So the numerator becomes (x - √(x² +1))/√(x² +1). Therefore, f'(x) = (x - √(x² +1)) / [ √(x² +1) * (√(x² +1) - x) * ln 2 ].Notice that (x - √(x² +1)) is negative because √(x² +1) > x for all real x. So x - √(x² +1) is negative. The denominator: √(x² +1) is positive, and (√(x² +1) - x) is positive, as before. So denominator is positive, numerator is negative, so overall f'(x) is negative. Therefore, f(x) is decreasing. So A is both odd and decreasing. So A is a candidate. Let me check other options too.Option B: f(x) = 1/x. First, check if it's odd. f(-x) = 1/(-x) = -1/x = -f(x), so yes, odd function. Now, is it decreasing? Let's check the derivative. f'(x) = -1/x². Since x² is always positive, f'(x) is always negative wherever it's defined (x ≠ 0). But wait, hold on. Is 1/x a decreasing function over its entire domain? The function 1/x is decreasing on each interval where it's defined, that is, on (-∞, 0) and (0, ∞). However, if you consider the entire domain, which is ℝ {0}, then it's not strictly decreasing overall because, for example, f(-1) = -1 and f(1) = 1. So even though it's decreasing on each separate interval, the function is not decreasing over the entire domain because of the asymptote at x=0. So depending on the interpretation, but the problem states "in its domain". The domain is all real numbers except 0, so technically, a function is decreasing if for any x1 < x2 in the domain, f(x1) > f(x2). However, since x1 and x2 have to be in the domain, but if x1 is negative and x2 is positive, even though x1 < x2, f(x1) is negative and f(x2) is positive, so f(x1) < f(x2). So actually, the function is not decreasing over its entire domain because of the jump from negative to positive over the asymptote. Therefore, even though it's decreasing on each separate interval, it's not decreasing on its entire domain. Therefore, option B is not a correct answer. Hmm, that's a point to note. So B might not be a correct answer.Moving on to option C: f(x) = x² - x³. First, check if it's odd. Let's compute f(-x) = (-x)² - (-x)³ = x² + x³. Compare to -f(x) = -x² + x³. So f(-x) ≠ -f(x) unless x² + x³ = -x² + x³, which would require 2x² = 0, which is only true for x=0. Therefore, not an odd function. So C is out.Option D: f(x) = sin x. Sin x is an odd function because sin(-x) = -sin x. So that's good. Now, is it a decreasing function? Let's check the derivative. f'(x) = cos x. The derivative of sin x is cos x, which is not always negative. Cos x is positive in (-π/2 + 2πk, π/2 + 2πk) and negative otherwise, where k is integer. Therefore, sin x is increasing where cos x is positive and decreasing where cos x is negative. So sin x is not a decreasing function over its entire domain. Hence, D is not correct.So, going back, A is odd and decreasing. B is odd but not decreasing over the entire domain. C is not odd. D is odd but not decreasing. So the answer should be A.Wait, let me just confirm again for option B. The problem says "decreasing function in its domain". If the domain is considered as two separate intervals, but the standard definition of a decreasing function requires that for any x1 < x2 in the entire domain, f(x1) ≥ f(x2). However, since in the case of 1/x, for x1 = -1 and x2 = 1, x1 < x2 but f(x1) = -1 < 1 = f(x2), which violates the decreasing condition. Therefore, 1/x is not decreasing on its entire domain, even though it is decreasing on each connected component of its domain. So B is invalid. Hence, A is the correct answer.Final Answerboxed{A}</think>To determine which function is both a decreasing function and an odd function, we analyze each option:Option A: ( f(x) = log_{2}(sqrt{x^{2}+1} - x) )1. Odd Function Check: [ f(-x) = log_{2}(sqrt{(-x)^{2}+1} - (-x)) = log_{2}(sqrt{x^{2}+1} + x) ] Using the property ((sqrt{x^{2}+1} - x)(sqrt{x^{2}+1} + x) = 1), we get: [ log_{2}(sqrt{x^{2}+1} + x) = log_{2}left(frac{1}{sqrt{x^{2}+1} - x}right) = -log_{2}(sqrt{x^{2}+1} - x) = -f(x) ] Hence, ( f(x) ) is odd.2. Decreasing Function Check: Compute the derivative: [ f'(x) = frac{d}{dx} left( log_{2}(sqrt{x^{2}+1} - x) right) ] Let ( u = sqrt{x^{2}+1} - x ). Then, [ frac{du}{dx} = frac{x}{sqrt{x^{2}+1}} - 1 = frac{x - sqrt{x^{2}+1}}{sqrt{x^{2}+1}} ] Since ( x - sqrt{x^{2}+1} < 0 ), the numerator is negative. The denominator ( (sqrt{x^{2}+1} - x) ln 2 ) is positive. Thus, ( f'(x) < 0 ), confirming ( f(x) ) is decreasing.Option B: ( f(x) = frac{1}{x} )1. Odd Function Check: [ f(-x) = frac{1}{-x} = -frac{1}{x} = -f(x) ] Hence, ( f(x) ) is odd.2. Decreasing Function Check: The derivative ( f'(x) = -frac{1}{x^{2}} ) is negative, but the function is not decreasing over its entire domain due to the discontinuity at ( x = 0 ).Option C: ( f(x) = x^{2} - x^{3} )1. Odd Function Check: [ f(-x) = (-x)^{2} - (-x)^{3} = x^{2} + x^{3} neq -f(x) ] Hence, ( f(x) ) is not odd.Option D: ( f(x) = sin x )1. Odd Function Check: [ f(-x) = sin(-x) = -sin x = -f(x) ] Hence, ( f(x) ) is odd.2. Decreasing Function Check: The derivative ( f'(x) = cos x ) is not always negative, so ( f(x) ) is not decreasing.Conclusion: The correct answer is (boxed{A}).,<think>Okay, so I have this problem here where I need to figure out which of the given functions is both a decreasing function and an odd function. The options are A, B, C, D. Let me start by recalling what each of those terms means.First, an odd function. I remember that an odd function satisfies the condition f(-x) = -f(x) for all x in its domain. So if I plug in -x into the function, it should be the same as negating the entire function. Classic examples are f(x) = x, f(x) = x^3, f(x) = sin(x), etc. So that's one part of the requirement.Then, the function also has to be a decreasing function. Now, a decreasing function is one where, as x increases, f(x) decreases. More formally, for any two points x1 and x2 in the domain, if x1 < x2, then f(x1) > f(x2). Alternatively, if the derivative f’(x) is negative for all x in the domain, then the function is decreasing everywhere. That might be a useful way to check.So, I need to check each of the options A, B, C, D to see if they are both odd and decreasing.Let me start with option B: f(x) = 1/x. Hmm, 1/x is a classic function. Let's check if it's odd. Let's compute f(-x) = 1/(-x) = -1/x = -f(x), so yes, it's odd. Good. Now, is it decreasing? Let's check its derivative. The derivative of 1/x is -1/x². Since x² is always positive (except at x=0, which isn't in the domain), the derivative is negative everywhere in the domain. So f(x) = 1/x is decreasing on both intervals (negative infinity to 0) and (0 to positive infinity). But wait, the function is decreasing on each interval separately, but is it decreasing over its entire domain? Hmm. Wait, the problem says "a decreasing function in its domain." The domain of 1/x is all real numbers except 0. However, a function can only be decreasing on an interval if it's defined on an interval. But here, the domain is split into two intervals. So, does the problem consider the function decreasing if it's decreasing on each separate interval? The standard definition requires the function to be decreasing on its entire domain, but since the domain here is not an interval but two separate intervals, technically, I think the function isn't decreasing on its entire domain. Because, for example, if you take x1 = -1 and x2 = 1, then x1 < x2, but f(x1) = -1 and f(x2) = 1, so f(x1) < f(x2), which would contradict being decreasing. But since the function isn't defined between -1 and 1 except at 0, which isn't included, maybe the standard is that it's decreasing on each connected component of its domain. But I need to check whether the problem's wording allows that. The problem says "a decreasing function in its domain". If the domain is considered as two separate intervals, then the function is decreasing on each of those intervals. But in the context of calculus, a function can only be considered decreasing on an interval. If the domain is not an interval, then the function can't be decreasing on its entire domain. So perhaps option B is not correct because it's not decreasing on its entire domain, only on each separate interval. So maybe B is not the answer. Let's hold that thought.Next, option D: f(x) = sin x. Sin x is an odd function, right? Because sin(-x) = -sin(x). So that's good. Is sin x a decreasing function? Well, let's think. The derivative of sin x is cos x. The derivative is cos x, which is positive in some intervals and negative in others. For example, cos x is positive in (-π/2, π/2) and negative in (π/2, 3π/2), etc. So sin x is not a decreasing function over its entire domain (which is all real numbers). It alternates between increasing and decreasing. Therefore, sin x is not a decreasing function everywhere. So D is out.Option C: f(x) = x² - x³. Let's check if this is an odd function. Let's compute f(-x) = (-x)^2 - (-x)^3 = x² - (-x³) = x² + x³. Then, -f(x) = -x² + x³. Comparing f(-x) and -f(x): f(-x) = x² + x³, while -f(x) = -x² + x³. These are not the same unless x² = 0, which is only at x=0. So f(-x) ≠ -f(x) in general, so C is not an odd function. Therefore, C is out.Now option A: f(x) = log₂(√(x² + 1) - x). Let's check if this is an odd function. Let's compute f(-x). Substitute -x into the function:f(-x) = log₂(√((-x)² + 1) - (-x)) = log₂(√(x² + 1) + x).Now, let's compare this to -f(x). Let's compute -f(x):-f(x) = -log₂(√(x² + 1) - x).We need to check if f(-x) = -f(x). So:log₂(√(x² + 1) + x) = -log₂(√(x² + 1) - x).Let me see if these two expressions are equal. Let's denote the argument of the logarithm on the left as A = √(x² +1) + x, and on the right, inside the log, we have B = √(x² +1) - x. Let's compute A * B:A * B = [√(x² +1) + x][√(x² +1) - x] = (√(x² +1))² - x² = (x² +1) - x² = 1.So A * B = 1. Therefore, A = 1/B. So, log₂(A) = log₂(1/B) = -log₂(B). Therefore, log₂(A) = -log₂(B). Which is exactly what we have: f(-x) = log₂(A) = -log₂(B) = -f(x). Therefore, f(-x) = -f(x). So A is an odd function. Good.Now, we need to check if A is a decreasing function. Let's compute the derivative of f(x) = log₂(√(x² +1) - x). First, let's simplify the expression inside the log. Maybe we can rewrite it.Let me note that √(x² +1) - x. Let me rationalize this expression. Multiply numerator and denominator by √(x² +1) + x. Wait, but since it's inside a logarithm, maybe that's not necessary. Alternatively, perhaps we can express this in a different way.Alternatively, let's compute the derivative. Let me recall that the derivative of log₂(u) is (1/(u ln 2)) * u'. So:f'(x) = [1 / ( (√(x² +1) - x ) * ln 2 ) ] * [ derivative of (√(x² +1) - x ) ]Compute the derivative inside:d/dx [√(x² +1)] = (1/(2√(x² +1))) * 2x = x / √(x² +1).Then, derivative of -x is -1. So the derivative is x / √(x² +1) - 1.Therefore, f'(x) = [ (x / √(x² +1) - 1 ) ] / [ (√(x² +1) - x ) * ln 2 ]Let me simplify the numerator and denominator.First, note that the numerator is (x / √(x² +1) - 1). Let's write it as:[ x - √(x² +1) ] / √(x² +1)So, numerator = [x - √(x² +1)] / √(x² +1)Denominator of the derivative is [√(x² +1) - x] * ln 2Notice that [x - √(x² +1)] = - [√(x² +1) - x]Therefore, numerator = - [√(x² +1) - x] / √(x² +1)Therefore, the derivative becomes:[ - [√(x² +1) - x] / √(x² +1) ] / [ (√(x² +1) - x ) * ln 2 ]Simplify numerator and denominator:The [√(x² +1) - x] terms cancel out (they are in both numerator and denominator), leaving:-1 / √(x² +1) / ln 2So f'(x) = -1 / [ √(x² +1) * ln 2 ]Since √(x² +1) is always positive, and ln 2 is a positive constant, the entire derivative f'(x) is negative for all real x. Therefore, the function f(x) is decreasing everywhere on its domain (which is all real numbers, since √(x² +1) - x is always positive. Let's check that: √(x² +1) > |x|, so √(x² +1) - x is positive when x is positive, because √(x² +1) - x = something bigger than x - x = 0. When x is negative, √(x² +1) - x is sqrt(positive) minus negative, so sqrt(sth) + |x|, which is also positive. Therefore, the expression inside the log is always positive, so the domain is all real numbers. Therefore, the derivative is always negative, so the function is decreasing on its entire domain. So option A is both odd and decreasing.Wait, but earlier with option B, we had f(x) = 1/x, which is odd, but is it decreasing? The problem is that the domain is split into two intervals, so the function is decreasing on each interval, but not on the entire domain. The question says "a decreasing function in its domain". If the problem considers each connected component separately, then maybe B is also acceptable. But in standard terminology, a function is decreasing on its domain only if the domain is an interval. If the domain is not an interval, then we say the function is decreasing on each interval of its domain. So maybe the problem is expecting the function to be decreasing on its entire domain, which for 1/x is not possible because the domain is two intervals.But let's confirm. Let's think about the function f(x) = 1/x. If you take x1 = -1 and x2 = 1, even though x1 < x2, f(x1) = -1 and f(x2) = 1, so f(x1) < f(x2). But since the function isn't defined between -1 and 1 (except at 0, which is excluded), the definition of a decreasing function requires that for any x1 < x2 in the domain, f(x1) > f(x2). However, since the domain is not connected, x1 and x2 can be in different intervals. Therefore, in such cases, the function is considered decreasing on each interval, but not necessarily on the entire domain. Therefore, if the question requires the function to be decreasing on its entire domain (i.e., for all x1 < x2 in the domain, f(x1) > f(x2)), then 1/x would not satisfy that because when you take x1 in (-infty, 0) and x2 in (0, infty), f(x1) is negative and f(x2) is positive, so f(x1) < f(x2), which violates the decreasing condition. Therefore, strictly speaking, 1/x is not a decreasing function on its entire domain. Therefore, option B is not correct. Therefore, option A is the correct answer.Therefore, the answer should be A.Final Answerboxed{A}</think>To determine which function is both a decreasing function and an odd function, we analyze each option:1. Odd Function Check: An odd function satisfies ( f(-x) = -f(x) ).2. Decreasing Function Check: A function is decreasing if its derivative is negative for all ( x ) in its domain.Option A: ( f(x) = log_{2}(sqrt{x^{2}+1} - x) )- Odd Function Check: [ f(-x) = log_{2}(sqrt{(-x)^{2} + 1} - (-x)) = log_{2}(sqrt{x^{2} + 1} + x) ] [ -f(x) = -log_{2}(sqrt{x^{2} + 1} - x) ] Using the property of logarithms and rationalizing the argument, we find ( f(-x) = -f(x) ), confirming it is odd.- Decreasing Function Check: [ f'(x) = frac{1}{(sqrt{x^{2} + 1} - x) ln 2} left( frac{x}{sqrt{x^{2} + 1}} - 1 right) ] Simplifying, we find the derivative is always negative, confirming it is decreasing.Option B: ( f(x) = frac{1}{x} )- Odd Function Check: [ f(-x) = frac{1}{-x} = -frac{1}{x} = -f(x) ] Confirming it is odd.- Decreasing Function Check: The derivative ( f'(x) = -frac{1}{x^{2}} ) is negative on each interval ( (-infty, 0) ) and ( (0, infty) ), but the function is not decreasing on the entire domain due to the split intervals.Option C: ( f(x) = x^{2} - x^{3} )- Odd Function Check: [ f(-x) = (-x)^{2} - (-x)^{3} = x^{2} + x^{3} neq -f(x) ] Confirming it is not odd.Option D: ( f(x) = sin x )- Odd Function Check: [ f(-x) = sin(-x) = -sin x = -f(x) ] Confirming it is odd.- Decreasing Function Check: The derivative ( f'(x) = cos x ) alternates in sign, so it is not decreasing everywhere.Conclusion: Option A is both an odd function and a decreasing function on its entire domain. boxed{A}]

🔑:Solution: A. The domain of f(x)=log_{2}(sqrt{x^{2}+1}-x) is mathbb{R}, and f(-x)=log_{2}(sqrt{x^{2}+1}+x)=log_{2}(sqrt{x^{2}+1}-x)^{-1}=-f(x), so f(x) is an odd function. Since y=sqrt{x^{2}+1}-x=frac{1}{sqrt{x^{2}+1}+x} is a decreasing function in its domain, the function f(x)=log_{2}(sqrt{x^{2}+1}-x) is a decreasing function in its domain, meeting the conditions; B. f(x)=frac{1}{x} is an odd function in the domain {x|xneq0}, and it is a decreasing function in (-infty,0) and (0,+infty), but it is not a decreasing function in its entire domain, not meeting the conditions; C. Since f(-x)=(-x)^{2}-(-x)^{3}=x^{2}+x^{3}, f(x) is neither odd nor even, not meeting the conditions; D. f(x)=sin x is an odd function, but it is not a monotonic function in its domain, not meeting the conditions. Therefore, the correct choice is: boxed{A}. This problem examines the judgment of the odd-even properties and monotonicity of functions, as well as the monotonicity of logarithmic functions and composite functions. Mastering the odd-even properties and monotonicity of common functions is key to solving the problem.

🧠:[<think>Okay, let's see. The problem is about finding an approximate root for the equation x³ + x - 1 = 0 using the Intermediate Value Theorem. The options given are intervals between 0.5 and 0.9, in increments of 0.1. So, the question is asking which interval the root must lie in based on evaluating the function at the endpoints and checking the sign changes.First, the problem statement mentions that since f(0) = -1 and f(1) = 1, there's a root between 0 and 1. Then, they checked f(0.5) and found it to be -0.375, which is still negative, so the root must be between 0.5 and 1. Now, we need to continue this process to narrow it down further to one of the given intervals: A (0.5, 0.6), B (0.6, 0.7), C (0.7, 0.8), or D (0.8, 0.9).Let me recall how the Intermediate Value Theorem works. If a function f is continuous on [a, b], and there's a value c between f(a) and f(b), then there exists some x in (a, b) such that f(x) = c. Since we're dealing with roots, we're looking for a sign change from negative to positive (or vice versa) which indicates that the function crossed zero in that interval.So, the strategy here is to compute the value of f(x) at the endpoints of each interval provided and see where the sign changes. Let me start by calculating f(0.5), which is already given as -0.375. Then, let's compute f(0.6), f(0.7), f(0.8), and f(0.9) to determine in which interval the root lies.Let me compute each step by step.First, f(0.5) = (0.5)³ + 0.5 - 1 = 0.125 + 0.5 - 1 = -0.375. That's correct as given.Now, let's check f(0.6):f(0.6) = (0.6)³ + 0.6 - 1 = 0.216 + 0.6 - 1 = 0.816 - 1 = -0.184.So, f(0.6) is still negative. That means the root is not in (0.5, 0.6), because both endpoints are negative. So interval A is out.Next, check f(0.7):f(0.7) = (0.7)³ + 0.7 - 1 = 0.343 + 0.7 - 1 = 1.043 - 1 = 0.043.Okay, so f(0.7) is approximately 0.043, which is positive. Therefore, since f(0.6) is -0.184 and f(0.7) is +0.043, the function crosses from negative to positive between 0.6 and 0.7. That suggests the root is in interval B: (0.6, 0.7). But wait, let's check the next intervals just to be thorough.Compute f(0.8):f(0.8) = (0.8)³ + 0.8 - 1 = 0.512 + 0.8 - 1 = 1.312 - 1 = 0.312.Positive again. So, between 0.7 and 0.8, f(x) is still positive. Similarly, f(0.9):f(0.9) = (0.9)³ + 0.9 - 1 = 0.729 + 0.9 - 1 = 1.629 - 1 = 0.629.Still positive. So, all intervals from 0.7 onwards have positive f(x) at their left endpoints. Since the root is where f(x) transitions from negative to positive, the first interval where the left endpoint is negative and the right is positive is (0.6, 0.7). Therefore, interval B should be the correct answer.But wait, let me verify once more. Maybe there's a miscalculation. Let me recalculate f(0.7):0.7 cubed is 0.7 * 0.7 = 0.49, then 0.49 * 0.7 = 0.343. Then, 0.343 + 0.7 = 1.043. Subtract 1, we get 0.043. That's correct.Similarly, f(0.6):0.6 * 0.6 = 0.36, 0.36 * 0.6 = 0.216. Then 0.216 + 0.6 = 0.816. Subtract 1, we get -0.184. Correct.So, between 0.6 and 0.7, the function goes from -0.184 to +0.043, so the root is in (0.6, 0.7). Therefore, option B.But wait, let me check the problem statement again. The problem says "since f(0.5) = -0.375 < 0, we know that the equation must have a root in the interval (0.5, 1); and so on, the interval in which this equation must have a root is ( )" So they have already narrowed it down to (0.5, 1) by checking f(0.5). Then, I assume the next step is to check the midpoint of 0.5 and 1, which is 0.75. But maybe the problem is expecting us to check the next intervals given in the options? Hmm.Wait, the problem says "and so on", so perhaps they are expecting to use the same method as before but with the given intervals. So maybe the problem is structured so that after establishing the root is in (0.5, 1), then check at 0.5, 0.6, 0.7, etc., each time subdividing into smaller intervals. But according to the previous step, they checked 0.5, found f(0.5) < 0, so the root is in (0.5, 1). Then, perhaps the next step is to check the midpoint of 0.5 and 1, which is 0.75, but f(0.75) is?Wait, let's compute f(0.75):0.75³ + 0.75 - 1. Let's calculate 0.75³: 0.75 * 0.75 = 0.5625, times 0.75 = 0.421875. Then add 0.75: 0.421875 + 0.75 = 1.171875. Subtract 1: 0.171875. So f(0.75) = 0.171875, which is positive. Therefore, since f(0.5) is -0.375 and f(0.75) is +0.171875, the root is in (0.5, 0.75). But the options given are (0.5, 0.6), (0.6, 0.7), (0.7, 0.8), etc. So perhaps they want a finer subdivision?Wait, perhaps instead of bisecting into 0.5 and 0.75, they are using increments of 0.1. So, starting from (0.5, 1), they check f(0.6), which is -0.184. Then, since f(0.6) is still negative, the root is in (0.6, 1). Then check f(0.7), which is 0.043. Positive, so the root is in (0.6, 0.7). Therefore, the answer is B. That seems to align with the previous steps. So each time, they check the next 0.1 increment.Alternatively, maybe the problem is expecting us to check the intervals given as options by evaluating f at their endpoints.So, for each interval:A: (0.5, 0.6). Compute f(0.5) = -0.375, f(0.6) = -0.184. Both negative, so no root here.B: (0.6, 0.7). f(0.6) = -0.184, f(0.7) = +0.043. Sign changes, so root here.C: (0.7, 0.8). f(0.7) = +0.043, f(0.8) = +0.312. Both positive, no root.D: (0.8, 0.9). Both positive as well. So the only interval with a sign change is B.Therefore, the answer should be B.Wait, but maybe I need to check if there's a mistake in my calculations?Let me verify f(0.6) again:0.6^3 is 0.6*0.6=0.36, 0.36*0.6=0.216. Then 0.216 + 0.6 -1 = 0.816 -1 = -0.184. Correct.f(0.7): 0.7^3=0.343, 0.343 +0.7 -1=1.043 -1=0.043. Correct.So, yes, the sign changes between 0.6 and 0.7. So answer is B.Alternatively, maybe the problem expects a different approach? For example, using Newton-Raphson or something else? But the problem statement says "using the following method", which refers to checking the sign at the endpoints, so the Intermediate Value Theorem approach.Therefore, the answer is B: (0.6, 0.7).Final Answerboxed{B}</think>To find the approximate root of the equation (x^3 + x - 1 = 0) using the Intermediate Value Theorem, we start by checking the function values at the endpoints of the given intervals.1. Interval (0.5, 0.6): - (f(0.5) = (0.5)^3 + 0.5 - 1 = -0.375) (negative) - (f(0.6) = (0.6)^3 + 0.6 - 1 = -0.184) (negative) - Both values are negative, so there is no root in this interval.2. Interval (0.6, 0.7): - (f(0.6) = -0.184) (negative) - (f(0.7) = (0.7)^3 + 0.7 - 1 = 0.043) (positive) - The function changes sign from negative to positive, indicating a root in this interval.3. Interval (0.7, 0.8): - (f(0.7) = 0.043) (positive) - (f(0.8) = (0.8)^3 + 0.8 - 1 = 0.312) (positive) - Both values are positive, so there is no root in this interval.4. Interval (0.8, 0.9): - (f(0.8) = 0.312) (positive) - (f(0.9) = (0.9)^3 + 0.9 - 1 = 0.629) (positive) - Both values are positive, so there is no root in this interval.Since the function (f(x)) changes sign from negative to positive in the interval ((0.6, 0.7)), the root must lie in this interval.(boxed{B}),<think>Okay, so I need to figure out which interval the root of the equation ( x^3 + x - 1 = 0 ) falls into. The options are A: (0.5, 0.6), B: (0.6, 0.7), C: (0.7, 0.8), D: (0.8, 0.9). Let me start by understanding the problem.First, the function is ( f(x) = x^3 + x - 1 ). They mentioned that since ( f(0) = -1 ) and ( f(1) = 1 ), there must be a root between 0 and 1 because the function crosses from negative to positive. Then, they checked ( f(0.5) = -0.375 ), which is still negative, so the root must be between 0.5 and 1. Now, we need to narrow it down further to the given intervals.I think this is an example of the bisection method, where you keep dividing the interval in half and checking the sign of the function at the midpoint to determine which subinterval contains the root. Let me try to apply that here.Starting with the interval (0.5, 1):1. First midpoint is 0.75. Let's calculate ( f(0.75) ). ( 0.75^3 + 0.75 - 1 = (0.421875) + 0.75 - 1 = 1.171875 - 1 = 0.171875 ). So ( f(0.75) = 0.171875 ), which is positive.2. Since ( f(0.5) = -0.375 ) and ( f(0.75) = 0.171875 ), the root is between 0.5 and 0.75. Now we check the midpoint of this new interval, which is 0.625.3. Calculate ( f(0.625) ): ( 0.625^3 + 0.625 - 1 ). Let's compute ( 0.625^3 ). 0.625 squared is 0.390625, multiplied by 0.625 gives 0.244140625. So, 0.244140625 + 0.625 - 1 = 0.869140625 - 1 = -0.130859375. So ( f(0.625) = -0.130859375 ), which is negative.4. Now we know the root is between 0.625 and 0.75. Next midpoint is 0.6875.5. Compute ( f(0.6875) ): 0.6875^3 + 0.6875 - 1. Let's calculate 0.6875^3. First, 0.6875 squared is 0.47265625. Multiply by 0.6875: 0.47265625 * 0.6875. Let me compute that step by step. 0.47265625 * 0.6 = 0.28359375, 0.47265625 * 0.08 = 0.0378125, and 0.47265625 * 0.0075 = approximately 0.003544921875. Adding those together: 0.28359375 + 0.0378125 = 0.32140625 + 0.003544921875 ≈ 0.324951171875. So, 0.324951171875 + 0.6875 - 1 ≈ 1.012451171875 - 1 = 0.012451171875. So ( f(0.6875) ≈ 0.01245 ), which is positive.6. Now, the interval is between 0.625 (negative) and 0.6875 (positive). Midpoint is (0.625 + 0.6875)/2 = 0.65625.7. Compute ( f(0.65625) ): ( 0.65625^3 + 0.65625 - 1 ). First, 0.65625 squared is 0.4306640625. Multiply by 0.65625: 0.4306640625 * 0.65625. Let me compute this. 0.4 * 0.65625 = 0.2625, 0.0306640625 * 0.65625 ≈ 0.020126953125. So total ≈ 0.2625 + 0.020126953125 ≈ 0.282626953125. Then, 0.282626953125 + 0.65625 - 1 ≈ 0.938876953125 - 1 ≈ -0.061123046875. So ( f(0.65625) ≈ -0.06112 ), which is negative.8. Now, the root is between 0.65625 and 0.6875. Midpoint is 0.671875.9. Compute ( f(0.671875) ): ( 0.671875^3 + 0.671875 - 1 ). Let's compute 0.671875 cubed. First, square it: 0.671875 * 0.671875. Let me compute that: 0.6 * 0.6 = 0.36 0.6 * 0.071875 = 0.043125 0.071875 * 0.6 = 0.043125 0.071875 * 0.071875 ≈ 0.005166015625 Adding up all the terms: 0.36 + 0.043125 + 0.043125 + 0.005166015625 ≈ 0.451416015625 So, squared is approximately 0.451416015625. Now multiply by 0.671875: 0.4 * 0.671875 = 0.26875 0.05 * 0.671875 = 0.03359375 0.001416015625 * 0.671875 ≈ 0.0009515380859375 Adding these: 0.26875 + 0.03359375 = 0.30234375 + 0.000951538 ≈ 0.3032952880859375 So, 0.3032952880859375 + 0.671875 - 1 ≈ 0.9751702880859375 - 1 ≈ -0.0248297119140625. So ( f(0.671875) ≈ -0.02483 ), negative.10. Therefore, the root is between 0.671875 and 0.6875. Next midpoint is 0.6796875.11. Compute ( f(0.6796875) ): Let's compute ( 0.6796875^3 + 0.6796875 - 1 ). First, compute ( 0.6796875^3 ). Let's square it first: 0.6796875 * 0.6796875. Let me compute that step by step. 0.6 * 0.6 = 0.36 0.6 * 0.0796875 = 0.0478125 0.0796875 * 0.6 = 0.0478125 0.0796875 * 0.0796875 ≈ 0.00635 Adding up: 0.36 + 0.0478125 + 0.0478125 + 0.00635 ≈ 0.461975 So, squared ≈ 0.461975. Now multiply by 0.6796875: 0.4 * 0.6796875 = 0.271875 0.06 * 0.6796875 = 0.04078125 0.001975 * 0.6796875 ≈ 0.001341796875 Adding those: 0.271875 + 0.04078125 = 0.31265625 + 0.001341796875 ≈ 0.313998046875 So, ( 0.6796875^3 ≈ 0.313998 ). Then, adding 0.6796875 gives 0.313998 + 0.6796875 ≈ 0.9936855. Subtract 1: 0.9936855 - 1 ≈ -0.0063145. So ( f(0.6796875) ≈ -0.0063 ), still negative.12. Now, the interval is between 0.6796875 and 0.6875. Midpoint is 0.68359375.13. Compute ( f(0.68359375) ): Let's compute ( 0.68359375^3 + 0.68359375 - 1 ). First, cube 0.68359375. Let's first square it: 0.68359375 * 0.68359375. Breaking it down: 0.6 * 0.6 = 0.36 0.6 * 0.08359375 = 0.05015625 0.08359375 * 0.6 = 0.05015625 0.08359375 * 0.08359375 ≈ 0.006988525390625 Adding up: 0.36 + 0.05015625 + 0.05015625 + 0.006988525 ≈ 0.467300025390625 Now multiply by 0.68359375: 0.4 * 0.68359375 = 0.2734375 0.06 * 0.68359375 = 0.041015625 0.007300025390625 * 0.68359375 ≈ 0.00499261474609375 Adding those: 0.2734375 + 0.041015625 = 0.314453125 + 0.004992615 ≈ 0.31944573974609375 So, cube ≈ 0.3194457. Then, add 0.68359375: 0.3194457 + 0.68359375 ≈ 1.00303945. Subtract 1: ≈ 0.00303945. So ( f(0.68359375) ≈ 0.00304 ), positive.14. Now, the interval is between 0.6796875 (where f is -0.0063) and 0.68359375 (where f is +0.00304). Midpoint is 0.681640625.15. Compute ( f(0.681640625) ): Let's approximate this. Maybe use previous calculations to save time. Since 0.6796875 gives -0.0063 and 0.68359375 gives +0.00304, the root is near 0.682. But let me compute it properly. Compute ( x = 0.681640625 ). Let's compute x^3: First, x^2: 0.681640625^2. Let's approximate. (0.68 + 0.001640625)^2 = 0.68^2 + 2*0.68*0.001640625 + (0.001640625)^2 ≈ 0.4624 + 0.00223125 + 0.0000027 ≈ 0.4624 + 0.00223125 ≈ 0.46463125 + 0.0000027 ≈ 0.46463395. Then, x^3 = x^2 * x ≈ 0.46463395 * 0.681640625. Let me compute that: 0.4 * 0.681640625 = 0.27265625 0.06 * 0.681640625 = 0.0408984375 0.00463395 * 0.681640625 ≈ 0.0031611328125 Adding those: 0.27265625 + 0.0408984375 = 0.3135546875 + 0.0031611328 ≈ 0.3167158203. Then, x^3 + x -1 ≈ 0.3167158203 + 0.681640625 - 1 ≈ 0.9983564453 - 1 ≈ -0.0016435547. So approximately -0.00164. So ( f(0.681640625) ≈ -0.00164 ), still negative.16. Now, the interval is between 0.681640625 and 0.68359375. Midpoint is 0.6826171875.17. Compute ( f(0.6826171875) ): Let's compute x^3 + x -1. x = 0.6826171875 x^2 = (0.6826171875)^2. Let me approximate: 0.68^2 = 0.4624 0.0026171875^2 ≈ 0.000006851 Cross term: 2 * 0.68 * 0.0026171875 ≈ 0.00357421875 So total x^2 ≈ 0.4624 + 0.00357421875 + 0.000006851 ≈ 0.4659810696. Then x^3 = x^2 * x ≈ 0.4659810696 * 0.6826171875. Let's compute: 0.4 * 0.6826171875 = 0.273046875 0.06 * 0.6826171875 = 0.04095703125 0.0059810696 * 0.6826171875 ≈ 0.0040859375 Adding those: 0.273046875 + 0.04095703125 = 0.31400390625 + 0.0040859375 ≈ 0.31808984375. Then, x^3 + x -1 = 0.31808984375 + 0.6826171875 -1 ≈ 1.00070703125 -1 ≈ 0.00070703125. So approximately 0.000707, positive.18. So now, the interval is between 0.681640625 (f ≈ -0.00164) and 0.6826171875 (f ≈ +0.000707). Midpoint is 0.68212890625.19. Compute ( f(0.68212890625) ): Let's do this quickly. Maybe average the previous two results. But better to compute. x = 0.68212890625 x^3 + x -1. Let's approximate: x ≈ 0.6821289 x^3 ≈ (0.6821289)^3. Let's compute x^2 first: x^2 ≈ 0.6821289 * 0.6821289 ≈ 0.4653 (exact calculation may differ, but for approximation, maybe 0.465). Then x^3 ≈ 0.465 * 0.6821289 ≈ 0.317. Then, x^3 + x -1 ≈ 0.317 + 0.6821289 -1 ≈ 0.9991289 -1 ≈ -0.000871. So approximately -0.00087. So f(0.68212890625) ≈ -0.00087, negative.20. Now, the interval is between 0.68212890625 (negative) and 0.6826171875 (positive). Midpoint is 0.682373046875.21. Compute ( f(0.682373046875) ): Let's approximate again. x = 0.682373046875 x^3 + x -1. x^3 ≈ (0.682373)^3. Let's compute x^2 first: x ≈ 0.682373 x^2 ≈ 0.682373 * 0.682373 ≈ 0.4653 (since (0.682)^2 ≈ 0.465, as before) x^3 ≈ 0.4653 * 0.682373 ≈ 0.3175 Then, x^3 + x -1 ≈ 0.3175 + 0.682373 ≈ 0.999873 -1 ≈ -0.000127. So approximately -0.000127.22. Very close to zero. Now, interval is between 0.682373046875 (approx -0.000127) and 0.6826171875 (approx +0.000707). Next midpoint is 0.6824951171875.23. Compute ( f(0.6824951171875) ): Let's approximate. x = 0.6824951171875 x^3 + x -1. x^3 ≈ (0.682495)^3. Let's do x^2 first: x^2 ≈ (0.682495)^2 ≈ 0.4653 x^3 ≈ 0.4653 * 0.682495 ≈ 0.3177 Then, x^3 + x -1 ≈ 0.3177 + 0.682495 ≈ 1.000195 -1 ≈ 0.000195. Positive.24. So now, the interval is between 0.682373046875 (≈-0.000127) and 0.6824951171875 (≈+0.000195). The root is approximately between 0.68237 and 0.68250. So, around 0.6824.But wait, the problem is asking which interval among A: (0.5, 0.6), B: (0.6, 0.7), C: (0.7, 0.8), D: (0.8, 0.9) contains the root. Wait, but according to the bisection steps above, the root is approximately 0.682, which is between 0.6 and 0.7, but the options are A: 0.5-0.6, B: 0.6-0.7, C: 0.7-0.8, D: 0.8-0.9. So 0.682 is in B: (0.6, 0.7). But wait, the original problem said "since f(0.5) = -0.375 < 0, we know that the equation must have a root in (0.5, 1); and so on, the interval in which this equation must have a root is ( )".But maybe the question is not expecting me to compute all the way to the precise approximation, but rather, check the function at the endpoints of the given intervals.Wait, let's check. The options are A: (0.5, 0.6), B: (0.6, 0.7), C: (0.7, 0.8), D: (0.8, 0.9). We already know the root is between 0.5 and 1. Then between 0.5 and 0.75. Then between 0.625 and 0.75. Then between 0.625 and 0.6875. Then between 0.65625 and 0.6875. Continuing until it narrows down.But perhaps the problem is expecting the student to check at the endpoints of the given intervals. Let's compute f(0.6), f(0.7), f(0.8), etc., to see in which interval the sign changes.Compute f(0.6): ( 0.6^3 + 0.6 -1 = 0.216 + 0.6 -1 = 0.816 -1 = -0.184 ). So f(0.6) = -0.184.Compute f(0.7): ( 0.7^3 + 0.7 -1 = 0.343 + 0.7 -1 = 1.043 -1 = 0.043 ). So f(0.7) = 0.043.Therefore, since f(0.6) is negative and f(0.7) is positive, by the Intermediate Value Theorem, there is a root in (0.6, 0.7). Therefore, the answer is B.Wait, that was straightforward. But why did I do all that bisection earlier? Maybe the problem is structured to lead the student through steps: first, check f(0)= -1, f(1)=1, so root in (0,1). Then f(0.5)= -0.375, so root in (0.5,1). Then maybe check f(0.75)=0.171875, so root in (0.5,0.75). Then maybe check midpoint of (0.5,0.75) which is 0.625, f(0.625)= -0.1308... So root in (0.625, 0.75). Then the next step is to check f(0.6875)≈0.01245, so root in (0.625,0.6875). Then, perhaps the question is structured to continue this process until the interval is narrowed down to one of the given options. But if we check the endpoints given in the options, like 0.6 and 0.7, then since f(0.6) is negative and f(0.7) is positive, the root is in (0.6,0.7), which is option B. So maybe this is the answer.But just to confirm, let me check. If I compute f(0.6) = -0.184 and f(0.7)=0.043, so yes, sign changes here. Therefore, the answer is B.I think the key here is that while the bisection steps were getting more precise, the question is giving specific intervals and we can check the function at the endpoints of those intervals. So since f(0.6) is negative and f(0.7) is positive, the root is between 0.6 and 0.7. So the answer is B.Final Answerboxed{B}</think>To determine the interval in which the root of the equation ( x^3 + x - 1 = 0 ) lies, we start by evaluating the function ( f(x) = x^3 + x - 1 ) at key points.1. We know ( f(0) = -1 ) and ( f(1) = 1 ), so there is a root in the interval ((0, 1)).2. Evaluating at ( x = 0.5 ), we get ( f(0.5) = -0.375 ), indicating the root is in ((0.5, 1)).3. Next, we check the midpoint ( x = 0.75 ), where ( f(0.75) = 0.171875 ), so the root is in ((0.5, 0.75)).4. Evaluating at ( x = 0.625 ), we get ( f(0.625) = -0.130859375 ), narrowing the interval to ((0.625, 0.75)).5. Checking ( x = 0.6875 ), we find ( f(0.6875) approx 0.01245 ), so the root is in ((0.625, 0.6875)).To determine the correct interval among the given options, we evaluate the function at the endpoints of the intervals provided:- ( f(0.6) = 0.6^3 + 0.6 - 1 = -0.184 )- ( f(0.7) = 0.7^3 + 0.7 - 1 = 0.043 )Since ( f(0.6) ) is negative and ( f(0.7) ) is positive, by the Intermediate Value Theorem, there must be a root in the interval ((0.6, 0.7)).Thus, the correct answer is (boxed{B}).]

🔑:Since f(0.6) = -0.184 0, they have opposite signs, according to the Intermediate Value Theorem, the equation must have a root in the interval (0.6, 0.7). Therefore, the correct choice is boxed{B}.

🧠:[<think>Okay, so I need to calculate the integral from 0 to 2 of (sqrt(4 - x²) - 2x) dx. The options are given as A through D, so I need to figure out which one is the correct answer. Let me break this down step by step.First, the integral is split into two parts: the integral of sqrt(4 - x²) dx minus the integral of 2x dx, both from 0 to 2. So I can handle these two integrals separately and then subtract the results. That seems manageable.Let me start with the first part: the integral of sqrt(4 - x²) dx from 0 to 2. Hmm, sqrt(4 - x²) looks like the equation of a semicircle. Specifically, the equation of a circle with radius 2 centered at the origin is x² + y² = 4, so solving for y gives y = sqrt(4 - x²), which is the upper semicircle. Therefore, the integral of sqrt(4 - x²) from 0 to 2 should represent the area under the curve from x=0 to x=2, which is a quarter of the circle. Since the radius is 2, the area of the full circle is πr² = π(2)² = 4π. A quarter of that would be π. So the integral from 0 to 2 of sqrt(4 - x²) dx is π. Wait, let me confirm that.Alternatively, I know that the integral of sqrt(a² - x²) dx from 0 to a is (π a²)/4. Since here a is 2, so (π * 2²)/4 = (4π)/4 = π. Yes, that's correct. So that part is π.Now the second integral is the integral from 0 to 2 of 2x dx. That's straightforward. The antiderivative of 2x is x², right? So evaluating from 0 to 2, it's (2)² - (0)² = 4 - 0 = 4. So that integral is 4.Therefore, putting it all together, the original integral is π - 4. So looking at the options, B is π - 4, which should be the answer. Let me check the options again:A: 2π -4B: π -4C: ln2 -4D: ln2 -2So yes, B is π -4, which matches my calculation. But wait, hold on, let me double-check the first integral. Because sometimes when integrating sqrt(a² - x²), you might use substitution or trigonometric substitution. Let me verify that with calculus instead of relying on the geometric interpretation.Let me compute ∫ sqrt(4 - x²) dx. Let's use substitution. Let x = 2 sinθ, so dx = 2 cosθ dθ. Then sqrt(4 - x²) becomes sqrt(4 - 4 sin²θ) = sqrt(4 cos²θ) = 2 cosθ. So the integral becomes ∫ 2 cosθ * 2 cosθ dθ = 4 ∫ cos²θ dθ. Then we can use the power-reduction formula: cos²θ = (1 + cos2θ)/2, so the integral becomes 4 * (1/2) ∫ (1 + cos2θ) dθ = 2 ∫1 dθ + 2 ∫ cos2θ dθ. Integrating term by term, the first integral is 2θ, and the second integral is 2*(1/2) sin2θ = sin2θ. So the total integral becomes 2θ + sin2θ + C.Now, reverting back to x. Since x = 2 sinθ, sinθ = x/2, so θ = arcsin(x/2). Also, sin2θ = 2 sinθ cosθ. Since sinθ = x/2, cosθ = sqrt(1 - (x²/4)) = sqrt(4 - x²)/2. Therefore, sin2θ = 2*(x/2)*(sqrt(4 - x²)/2) = x*sqrt(4 - x²)/2.Putting it all together, the antiderivative is 2θ + sin2θ + C = 2 arcsin(x/2) + (x sqrt(4 - x²))/2 + C.Now, evaluating from 0 to 2. At x=2: arcsin(2/2) = arcsin(1) = π/2. So 2*(π/2) = π. The second term is (2 * sqrt(4 - 4))/2 = (2*0)/2 = 0. So the upper limit is π + 0 = π.At x=0: arcsin(0) = 0. The second term is (0 * sqrt(4 - 0))/2 = 0. So the lower limit is 0 + 0 = 0.Therefore, the definite integral from 0 to 2 of sqrt(4 - x²) dx is π - 0 = π. So that checks out with the geometric interpretation. Good.And the integral of 2x from 0 to 2 is indeed 4, as calculated earlier. So the entire original integral is π - 4, which is option B.Wait a second, but let me check the answer options again. A is 2π -4. Could I have made a mistake in the area interpretation? For example, maybe the integral from 0 to 2 of sqrt(4 - x²) is not a quarter circle?Wait, the full circle is x from -2 to 2, y from 0 to sqrt(4 - x²) is the upper half, so the area would be (1/2)πr² = (1/2)π(4) = 2π. But integrating from 0 to 2 would be a quarter of the circle? Wait, no. If you consider the entire circle, integrating from -2 to 2 gives the area of the upper half, which is 2π. But if you integrate from 0 to 2, that's a quarter of the circle? Wait, actually, the upper half of the circle is from x=-2 to x=2, y=0 to y=sqrt(4 -x²). So if we take x from 0 to 2, that's half of the upper half, which is a quarter of the full circle. So area is (1/4)π(2)^2 = π. That's correct. So yes, the integral is π. So the answer is π -4, which is option B. Therefore, the correct answer is B.But just to be thorough, let me compute the integral step by step without relying on geometric interpretation.So, as we did earlier, substitution x = 2 sinθ, which led us to the antiderivative 2 arcsin(x/2) + (x sqrt(4 -x²))/2. Evaluated from 0 to 2 gives π -0 = π. So that's correct. Then 2x integral is 4. So total is π -4. So answer is B.Therefore, I think the answer is B.Final Answerboxed{B}</think>To calculate the integral (int_{0}^{2} (sqrt{4-x^2} - 2x) , dx), we can break it into two separate integrals:1. The integral of (sqrt{4 - x^2}) from 0 to 2.2. The integral of (2x) from 0 to 2.For the first integral, (int_{0}^{2} sqrt{4 - x^2} , dx), we recognize that this represents the area of a quarter-circle with radius 2. The area of a full circle is (pi times 2^2 = 4pi), so a quarter-circle has area (pi). Thus, (int_{0}^{2} sqrt{4 - x^2} , dx = pi).For the second integral, (int_{0}^{2} 2x , dx), we find the antiderivative of (2x), which is (x^2). Evaluating from 0 to 2 gives:[left[ x^2 right]_{0}^{2} = 2^2 - 0^2 = 4 - 0 = 4]Subtracting the second integral from the first, we get:[pi - 4]Thus, the correct answer is (boxed{B}).,<think>Okay, so I need to solve this integral from 0 to 2 of (sqrt(4 - x^2) - 2x) dx. The answer choices are given, so maybe I can check them after I compute the integral. Let me break this down into two separate integrals because the integral of a difference is the difference of the integrals. That should make it easier.First, let's consider the integral from 0 to 2 of sqrt(4 - x^2) dx. Hmm, I remember that the integral of sqrt(a^2 - x^2) dx is something related to the area of a circle, right? Because sqrt(a^2 - x^2) is the equation for the upper half of a circle with radius a. So if I integrate that from 0 to a, it's like finding the area of a quarter of a circle. So in this case, a is 2, since we have sqrt(4 - x^2). Therefore, the integral from 0 to 2 of sqrt(4 - x^2) dx should be (1/4) of the area of a circle with radius 2. The area of a circle is πr², so π*(2)² = 4π. Then a quarter of that would be π. Wait, but is it a quarter?Let me think. The full circle would be from -2 to 2 in x and y. If we're integrating from 0 to 2, and the function is the upper half, then yes, that's a quarter of the circle. So the integral from 0 to 2 of sqrt(4 - x^2) dx is (1/4)*4π = π. So that part should be π.Okay, that seems straightforward. Now the second integral is from 0 to 2 of 2x dx. That's simpler. The integral of 2x is x², so evaluated from 0 to 2, it's (2)² - (0)² = 4 - 0 = 4. But since we have the integral of (sqrt(4 - x^2) - 2x) dx, that would be the first integral minus the second integral. So π - 4.Looking at the answer choices, option B is π - 4. So that must be the correct answer. But let me double-check to make sure I didn't make any mistakes.Wait, let me verify the first integral again. The integral of sqrt(a² - x²) dx from 0 to a is indeed (π a²)/4. So with a = 2, it's (π*4)/4 = π. Yep, that's correct. And the integral of 2x from 0 to 2 is 4. So subtracting gives π - 4. That's option B. Hmm, the options are labeled A: 2π -4, B: π -4, so yeah, B is the one.Wait, but just to be thorough, maybe I should compute the first integral using substitution to be sure. Let's try integrating sqrt(4 - x^2) dx. Let me set x = 2 sinθ. Then dx = 2 cosθ dθ. When x = 0, θ = 0. When x = 2, θ = π/2. So substituting, the integral becomes sqrt(4 - 4 sin²θ) * 2 cosθ dθ from 0 to π/2. Simplify inside the square root: sqrt(4(1 - sin²θ)) = sqrt(4 cos²θ) = 2 cosθ. So the integral becomes 2 cosθ * 2 cosθ dθ = 4 cos²θ dθ from 0 to π/2.Then, using the power-reduction formula: cos²θ = (1 + cos2θ)/2. So the integral becomes 4*(1 + cos2θ)/2 dθ = 2*(1 + cos2θ) dθ. Integrating that from 0 to π/2 gives 2 [θ + (sin2θ)/2] evaluated from 0 to π/2. At π/2: 2 [π/2 + (sinπ)/2] = 2*(π/2 + 0) = π. At 0: 2 [0 + 0] = 0. So the integral is π - 0 = π. Yep, that's the same result as before. So that's correct.So the first integral is π, the second is 4, so the answer is π - 4. Option B. Hmm, seems right.Alternatively, maybe graph both functions and see the area between them? The sqrt(4 - x^2) is a semicircle, but since we're integrating from 0 to 2, it's a quarter-circle in the first quadrant. Then the line y = 2x is a straight line starting at (0,0) and going to (2,4). So the integral is the area under the semicircle minus the area under the line from 0 to 2. The area under the semicircle is π, as established, and the area under the line is a triangle with base 2 and height 4, which would be (1/2)*2*4=4. So the difference is π - 4. That also makes sense.So yeah, the answer is definitely B: π -4.Final Answerboxed{B}</think>To calculate the integral (int_{0}^{2} (sqrt{4-x^2} - 2x) , dx), we break it into two separate integrals:1. The integral of (sqrt{4 - x^2}) from 0 to 2: - The function (sqrt{4 - x^2}) represents the upper half of a circle with radius 2. The integral from 0 to 2 corresponds to the area of a quarter-circle. - The area of a full circle with radius 2 is (4pi), so a quarter-circle has area (pi). - Therefore, (int_{0}^{2} sqrt{4 - x^2} , dx = pi).2. The integral of (2x) from 0 to 2: - The integral of (2x) is (x^2). - Evaluating from 0 to 2 gives (2^2 - 0^2 = 4). - Therefore, (int_{0}^{2} 2x , dx = 4).Subtracting the second integral from the first gives:[int_{0}^{2} (sqrt{4 - x^2} - 2x) , dx = pi - 4]Thus, the correct answer is (boxed{B}).]

🔑:To solve, int_{0}^{2} (sqrt{4-x^2}-2x)dx= int_{0}^{2} sqrt{4-x^2}dx+ int_{0}^{2}(-2x)dx, The geometric meaning of int_{0}^{2} sqrt{4-x^2}dx represents a quarter of the area of a circle with radius 2 centered at the origin, therefore int_{0}^{2} sqrt{4-x^2}dx= dfrac{1}{4}timespi r^2=pi, int_{0}^{2}(-2x)dx=-x^2 bigg|_{0}^{2}=-4, therefore int_{0}^{2} (sqrt{4-x^2}-2x)dx= int_{0}^{2} sqrt{4-x^2}dx+ int_{0}^{2}(-2x)dx=pi-4, therefore int_{0}^{2} (sqrt{4-x^2}-2x)dx=pi-4, Thus, the correct option is boxed{B}. Based on the operation of definite integrals, int_{0}^{2} (sqrt{4-x^2}-2x)dx= int_{0}^{2} sqrt{4-x^2}dx+ int_{0}^{2}(-2x)dx, and understanding the geometric meaning of definite integrals, the answer can be obtained. This question tests the operation of definite integrals, the geometric meaning of definite integrals, and computational skills, and is considered a medium-level problem.

🧠:[<think>Okay, let me try to solve these equations one by one. I'll start with the first one: (x^2 - 6x - 2 = 0). Hmm, this is a quadratic equation. I remember there are a few methods to solve quadratic equations, like factoring, completing the square, or using the quadratic formula. Let me check if this can be factored. The coefficients are 1, -6, and -2. So, I need two numbers that multiply to -2 (the constant term) and add up to -6 (the coefficient of the x term). Wait, let's see... factors of -2 could be 1 and -2 or -1 and 2. But 1 + (-2) = -1 and -1 + 2 = 1, neither of which is -6. So factoring doesn't seem to work here. Alright, moving on to the quadratic formula then. The quadratic formula is (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}). For the equation (x^2 - 6x - 2 = 0), the coefficients are a = 1, b = -6, c = -2. Plugging these into the formula, we get:(x = frac{-(-6) pm sqrt{(-6)^2 - 4(1)(-2)}}{2(1)})Simplifying that:(x = frac{6 pm sqrt{36 + 8}}{2})Because ( -4ac = -4*1*(-2) = +8), so 36 + 8 is 44. Therefore:(x = frac{6 pm sqrt{44}}{2})Now, (sqrt{44}) can be simplified. Since 44 is 4*11, so (sqrt{44} = sqrt{4*11} = 2sqrt{11}). Therefore, substituting back in:(x = frac{6 pm 2sqrt{11}}{2})We can factor out a 2 in the numerator:(x = frac{2(3 pm sqrt{11})}{2})Which simplifies to:(x = 3 pm sqrt{11})So, the solutions to the first equation are (x = 3 + sqrt{11}) and (x = 3 - sqrt{11}). Let me just check if that makes sense. If I plug x = 3 + sqrt(11) into the original equation:( (3 + sqrt{11})^2 - 6(3 + sqrt{11}) - 2 )Calculating the square:= 9 + 6sqrt{11} + 11 - 18 - 6sqrt{11} - 2Combine like terms:9 + 11 - 18 - 2 = (20 - 20) = 0And the middle terms: 6sqrt{11} - 6sqrt{11} = 0. So yes, it works. Similarly, the other solution should also work. Good.Now, moving on to the second equation: ((2x + 1)^2 = -6x - 3). Let's see. First, I need to expand the left side and then rearrange the equation to form a quadratic equation. Let's expand the left side:Left side: ((2x + 1)^2 = (2x)^2 + 2*2x*1 + 1^2 = 4x^2 + 4x + 1)So the equation becomes:(4x^2 + 4x + 1 = -6x - 3)Now, let's bring all terms to one side to set the equation to zero. Let's add 6x and add 3 to both sides:Left side: 4x^2 + 4x + 1 + 6x + 3Right side: 0Combining like terms:4x^2 + (4x + 6x) + (1 + 3) = 4x^2 + 10x + 4So the equation is now:(4x^2 + 10x + 4 = 0)Wait, is that right? Let me check:Original equation: (2x+1)^2 = -6x -3Left side: 4x² +4x +1Right side: -6x -3Bring all terms to left: 4x² +4x +1 +6x +3 = 04x² + (4x +6x) + (1 +3) = 4x² +10x +4 =0. Yes, that's correct.So now, we have the quadratic equation 4x² +10x +4=0. Let's try to simplify this. All coefficients are even numbers, so maybe we can divide the entire equation by 2 to make it simpler:(4x² +10x +4)/2 = 0/2 => 2x² +5x +2=0So the simplified equation is 2x² +5x +2=0. Let's see if this can be factored. Looking for two numbers that multiply to 2*2=4 and add up to 5. Hmm, factors of 4 are 1 & 4, 2 & 2. 1 and 4 add up to 5. So yes:2x² +5x +2 = 2x² +4x +x +2 = (2x² +4x) + (x +2) = 2x(x +2) +1(x +2) = (2x +1)(x +2)Therefore, factored form is (2x +1)(x +2)=0. So the solutions are when 2x +1=0 or x +2=0. Therefore:2x +1=0 => x= -1/2x +2=0 => x= -2So the solutions are x= -1/2 and x= -2. Let me check these in the original equation to verify.First, x = -1/2:Left side: (2*(-1/2) +1)^2 = (-1 +1)^2 = 0^2 =0Right side: -6*(-1/2) -3 = 3 -3=0. So 0=0. Correct.Second, x= -2:Left side: (2*(-2) +1)^2 = (-4 +1)^2 = (-3)^2=9Right side: -6*(-2) -3 =12 -3=9. So 9=9. Correct.Therefore, both solutions work.But wait, hold on. The original equation was (2x +1)^2 = -6x -3. Let me check if the quadratic equation was formed correctly. Because when moving terms, sometimes people make sign errors.Original: (2x +1)^2 = -6x -3Left side after expansion: 4x² +4x +1Set equal to -6x -3. Then bringing all to left:4x² +4x +1 +6x +3=0Which is 4x² +10x +4=0. Then dividing by 2: 2x² +5x +2=0. Factored to (2x+1)(x+2)=0. So solutions x=-1/2 and x=-2. Both check out. So that's correct.Alternatively, could we have approached the second equation differently without expanding?Let me think. The original equation is (2x +1)^2 = -6x -3. Maybe notice that the right side is -3(2x +1). Wait, let's check:-6x -3 = -3(2x +1). Yes, because -3*2x = -6x and -3*1 = -3. So the equation can be rewritten as:(2x +1)^2 = -3(2x +1)Then, bring all terms to left:(2x +1)^2 +3(2x +1) =0Factor out (2x +1):(2x +1)(2x +1 +3)=0So, (2x +1)(2x +4)=0Therefore, solutions are:2x +1=0 => x= -1/22x +4=0 => x= -2Same solutions. So this alternative method works as well. That was clever, factoring out (2x +1). That might have saved some steps. But either way, the solutions are the same.So, both equations are solved. The first one has two real irrational solutions, and the second one has two real rational solutions.Final AnswerThe solutions are:(1) (boxed{3 + sqrt{11}}) and (boxed{3 - sqrt{11}});(2) (boxed{-dfrac{1}{2}}) and (boxed{-2}).</think>To solve the equations:(1) (x^2 - 6x - 2 = 0)We use the quadratic formula (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}) where (a = 1), (b = -6), and (c = -2):[x = frac{-(-6) pm sqrt{(-6)^2 - 4 cdot 1 cdot (-2)}}{2 cdot 1} = frac{6 pm sqrt{36 + 8}}{2} = frac{6 pm sqrt{44}}{2} = frac{6 pm 2sqrt{11}}{2} = 3 pm sqrt{11}]The solutions are (boxed{3 + sqrt{11}}) and (boxed{3 - sqrt{11}}).(2) ((2x + 1)^2 = -6x - 3)First, expand and rearrange the equation:[(2x + 1)^2 = -6x - 3 implies 4x^2 + 4x + 1 = -6x - 3 implies 4x^2 + 10x + 4 = 0]Divide by 2 to simplify:[2x^2 + 5x + 2 = 0]Factor the quadratic equation:[(2x + 1)(x + 2) = 0]The solutions are:[2x + 1 = 0 implies x = -frac{1}{2} quad text{and} quad x + 2 = 0 implies x = -2]The solutions are (boxed{-dfrac{1}{2}}) and (boxed{-2}).,<think>Okay, so I have two equations to solve here. Let's start with the first one: (x^2 - 6x - 2 = 0). Hmm, this looks like a quadratic equation. Quadratic equations can usually be solved by factoring, completing the square, or using the quadratic formula. Let me see if this one factors easily. The coefficients are 1, -6, and -2. To factor it, I need two numbers that multiply to -2 and add up to -6. But wait, -2 is the constant term, so actually, the numbers should multiply to (1)(-2) = -2 and add to -6. That doesn't seem possible with integers. Let me check: 1 and -2 multiply to -2 but add to -1. If I try other combinations, maybe with fractions? But that's complicated. Maybe factoring isn't the way to go here. Alright, let's try completing the square then. The general method for completing the square is to take the coefficient of x, divide by 2, square it, and add/subtract as needed. So for (x^2 - 6x - 2 = 0), the coefficient of x is -6. Half of that is -3, and squaring it gives 9. So I can rewrite the equation as:(x^2 - 6x + 9 - 9 - 2 = 0)Which simplifies to:((x - 3)^2 - 11 = 0)Then, moving the constant term to the other side:((x - 3)^2 = 11)Taking the square root of both sides:(x - 3 = pmsqrt{11})Therefore, solving for x gives:(x = 3 pm sqrt{11})So the solutions are (3 + sqrt{11}) and (3 - sqrt{11}). That seems right. Let me check by substituting back into the original equation. Take (x = 3 + sqrt{11}):First term: ((3 + sqrt{11})^2 = 9 + 6sqrt{11} + 11 = 20 + 6sqrt{11})Second term: -6x = -6*(3 + sqrt{11}) = -18 - 6sqrt{11}Third term: -2Adding all together: 20 + 6sqrt{11} -18 -6sqrt{11} -2 = (20 -18 -2) + (6sqrt{11} -6sqrt{11}) = 0 + 0 = 0. Perfect, that works. Similarly, for (x = 3 - sqrt{11}), it should also work. So that's the first equation done.Moving on to the second equation: ((2x + 1)^2 = -6x - 3). Hmm, let's see. First, maybe I should expand the left side and bring all terms to one side. Let me do that.Expanding the left side:((2x + 1)^2 = 4x^2 + 4x + 1)So the equation becomes:4x^2 + 4x + 1 = -6x -3Now, let's bring all terms to the left side:4x^2 + 4x + 1 + 6x + 3 = 0Combine like terms:4x^2 + (4x + 6x) + (1 + 3) = 0Which is:4x^2 + 10x + 4 = 0So we have a quadratic equation: 4x^2 + 10x + 4 = 0. Let's see if we can simplify this. All coefficients are even, so maybe divide by 2:2x^2 + 5x + 2 = 0Now, let's try factoring this. The quadratic is 2x^2 +5x +2. To factor, we need two numbers that multiply to (2)(2)=4 and add to 5. Wait, actually, for factoring with a leading coefficient, the method is a bit different. Let me think. Looking for two numbers that multiply to 2*2=4 and add to 5. Hmm, 4 and 1: 4*1=4 and 4+1=5. Yes! So we can split the middle term:2x^2 + 4x + x + 2 = 0Group the terms:(2x^2 + 4x) + (x + 2) = 0Factor out common terms:2x(x + 2) + 1(x + 2) = 0Now factor out (x + 2):(2x + 1)(x + 2) = 0Setting each factor equal to zero:2x + 1 = 0 => x = -1/2x + 2 = 0 => x = -2So the solutions are x = -1/2 and x = -2. Let me check these in the original equation.First, x = -1/2:Left side: (2*(-1/2) + 1)^2 = (-1 + 1)^2 = (0)^2 = 0Right side: -6*(-1/2) -3 = 3 -3 = 0So 0 = 0, correct.Next, x = -2:Left side: (2*(-2) +1)^2 = (-4 +1)^2 = (-3)^2 = 9Right side: -6*(-2) -3 = 12 -3 = 99 = 9, correct.So both solutions are valid. Therefore, the solutions are x = -1/2 and x = -2.But wait a second, the original equation was ((2x + 1)^2 = -6x -3). Let me just make sure that when we squared the left side, we didn't introduce any extraneous solutions. However, since both solutions check out, I think they are both valid.Alternatively, could there be another approach? Maybe expanding wasn't necessary. Let's see:Original equation: ((2x + 1)^2 = -6x -3)Take square roots? Hmm, but the right side is -6x -3. If I take square roots, I have to consider that the left side is a square, so it's non-negative, which implies the right side must also be non-negative. So perhaps before solving, we can note that -6x -3 ≥ 0. Let's see:-6x -3 ≥ 0 => -6x ≥ 3 => x ≤ -0.5So any solutions must satisfy x ≤ -0.5. Our solutions are x = -1/2 and x = -2, which both satisfy x ≤ -0.5. So that's consistent. Therefore, no extraneous solutions here.Alternatively, could we have solved this equation without expanding? Let's see:((2x +1)^2 = -6x -3)Let me set y = 2x +1. Then the equation becomes y^2 = -6x -3. But since y = 2x +1, we can express x in terms of y: x = (y -1)/2. Substitute back into the equation:y^2 = -6*( (y -1)/2 ) -3Simplify:y^2 = -3*(y -1) -3y^2 = -3y +3 -3y^2 = -3yBring all terms to one side:y^2 + 3y = 0Factor:y(y + 3) = 0So y = 0 or y = -3Then, substitute back for x:If y = 0: 0 = 2x +1 => x = -1/2If y = -3: -3 = 2x +1 => 2x = -4 => x = -2Same solutions. So that's another way. Either way, the solutions are x = -1/2 and x = -2.So both equations solved. Let me recap:Equation (1): Solutions are x = 3 + sqrt(11) and x = 3 - sqrt(11).Equation (2): Solutions are x = -1/2 and x = -2.Everything checks out. I don't think I made any mistakes here. Both equations are quadratic, and in each case, we solved them appropriately either by completing the square or factoring, and verified the solutions.Final AnswerThe solutions are (boxed{3 + sqrt{11}}), (boxed{3 - sqrt{11}}) for equation (1), and (boxed{-dfrac{1}{2}}), (boxed{-2}) for equation (2).</think>To solve the first equation (x^2 - 6x - 2 = 0), we use the method of completing the square:1. Rewrite the equation by completing the square: [ x^2 - 6x - 2 = 0 implies (x - 3)^2 - 11 = 0 ]2. Solve for (x) by taking the square root of both sides: [ (x - 3)^2 = 11 implies x - 3 = pmsqrt{11} implies x = 3 pm sqrt{11} ]The solutions to the first equation are (boxed{3 + sqrt{11}}) and (boxed{3 - sqrt{11}}).For the second equation ((2x + 1)^2 = -6x - 3):1. Expand and simplify the equation: [ (2x + 1)^2 = -6x - 3 implies 4x^2 + 4x + 1 = -6x - 3 implies 4x^2 + 10x + 4 = 0 ]2. Divide by 2 to simplify: [ 2x^2 + 5x + 2 = 0 ]3. Factor the quadratic equation: [ (2x + 1)(x + 2) = 0 ]4. Solve for (x): [ 2x + 1 = 0 implies x = -frac{1}{2} quad text{or} quad x + 2 = 0 implies x = -2 ]The solutions to the second equation are (boxed{-dfrac{1}{2}}) and (boxed{-2}).]

🔑:Solution: Solve: (1) From the original equation, we get x^2-6x=2, Completing the square, we obtain x^2-6x+3^2=2+3^2, Therefore, (x-3)^2=11, Taking the square root directly, we get x-3=pmsqrt{11}, Solving for x, we find x_1=3+sqrt{11}, x_2=3-sqrt{11}; (2) From the original equation, we get (2x+1)(2x+4)=0, Therefore, 2x+1=0 or 2x+4=0, Solving for x, we find x_1=-frac{1}{2}, x_2=-2.Thus, the solutions are x_1=3+sqrt{11}, x_2=3-sqrt{11} for the first equation, and x_1=-frac{1}{2}, x_2=-2 for the second equation. Therefore, the final answers are boxed{x_1=3+sqrt{11}, x_2=3-sqrt{11}} for the first equation and boxed{x_1=-frac{1}{2}, x_2=-2} for the second equation.