🧠:[<think>Okay, so the problem is about a student who gets five points for each correct answer and loses three points for each incorrect or unsolved problem. The teacher gave 80 problems, and the student ended up with eight points. We need to find out how many problems the student solved correctly. Hmm, let's see. First, let's break down the information. Total problems are 80. For each correct one, you gain 5 points, and for each incorrect or unsolved, you lose 3 points. The total score is 8. So, we need to find the number of correct answers. Let me define variables. Let's let x be the number of correct answers. Then, since there are 80 problems in total, the number of incorrect or unsolved problems would be 80 minus x, right? Because whatever isn't correct is either incorrect or not solved. Now, the total points come from the correct answers minus the points lost from the incorrect ones. So, for each correct answer, it's +5, so total points from correct answers would be 5x. Then, for each incorrect answer, it's -3, so total points lost would be 3*(80 - x). But since it's a deduction, the total points would be 5x - 3*(80 - x). And the result of that calculation is 8 points. So, we can set up the equation: 5x - 3(80 - x) = 8Okay, let's solve that step by step. First, expand the terms. Multiply the -3 into the parentheses. So:5x - 3*80 + 3x = 8Wait, hold on. Wait, the equation is 5x minus 3 times (80 - x). So that would be 5x - 240 + 3x? Wait, no. Wait, distributing the -3 over (80 - x) would be (-3)*80 + (-3)*(-x), which is -240 + 3x. So, combining like terms:5x + 3x - 240 = 8So 8x - 240 = 8Then, add 240 to both sides:8x = 8 + 2408x = 248Then divide both sides by 8:x = 248 / 8Let me do that division. 8 goes into 24 three times (3*8=24), remainder 0. Then bring down the 8. 8 goes into 8 once. So 31. Wait, 8*31 is 248. So x = 31. Wait, so the student solved 31 problems correctly. Let me check if that's right. So, correct answers: 31, incorrect: 80 - 31 = 49. Total points: 31*5 = 155. Points lost: 49*3 = 147. So net points: 155 - 147 = 8. Yep, that matches. So 31 is the right answer. But let me make sure I didn't make any miscalculations. Let me recalculate the equation. Starting with 5x - 3(80 - x) = 8Expand: 5x - 240 + 3x = 8Combine like terms: 8x - 240 = 8Add 240: 8x = 248Divide by 8: x = 31. Correct. Alternatively, maybe there's another way to approach it, like assuming all problems were incorrect first and then seeing how many need to be converted to correct. Let's try that for confirmation. If all 80 problems were incorrect, the student would lose 80*3 = 240 points. But they ended up with 8 points. So the difference between -240 and +8 is 248 points. Each correct answer converts a -3 into a +5, so the difference per correct answer is 5 + 3 = 8 points. So to gain 248 points, the number of correct answers needed is 248 / 8 = 31. Yep, same answer. So, that confirms it. So the student solved 31 problems correctly. Therefore, the answer is 31.Final AnswerThe student solved boxed{31} problems correctly.</think>Let ( x ) be the number of problems solved correctly. The number of problems not solved or solved incorrectly is ( 80 - x ). The total points from correct answers are ( 5x ) and the points deducted for incorrect answers are ( 3(80 - x) ). The equation for the total points is:[5x - 3(80 - x) = 8]Expanding and simplifying the equation:[5x - 240 + 3x = 8]Combining like terms:[8x - 240 = 8]Adding 240 to both sides:[8x = 248]Dividing by 8:[x = frac{248}{8} = 31]Verification: Correct answers = 31, incorrect answers = 49. Total points: ( 31 times 5 = 155 ). Points lost: ( 49 times 3 = 147 ). Net points: ( 155 - 147 = 8 ), which matches the given total.Thus, the student solved boxed{31} problems correctly.,<think>Okay, let's see. So the problem is about a student who attempted 80 problems. For each correct one, he gets 5 points, and for each incorrect or unsolved one, they deduct 3 points. In the end, he has 8 points. The question is how many problems did he solve correctly?Hmm. Alright, let me break this down. Let me start by trying to figure out what variables I need here. I think I need to define two variables: the number of correct answers and the number of incorrect answers. Let's say the number of correct answers is "c", and the number of incorrect or unsolved ones is "i". Since there are 80 problems total, that should give me the first equation:c + i = 80That seems straightforward. Now, the next part is about the points. For each correct answer, he gets 5 points, so total points from correct answers would be 5c. For each incorrect answer, they deduct 3 points, so that's -3i. The total points he ends up with is 8. So combining these, the second equation should be:5c - 3i = 8Now, so we have a system of two equations:1. c + i = 802. 5c - 3i = 8I need to solve this system to find the value of c. Let me think about how to do that. The standard way is to solve one of the equations for one variable and substitute into the other. Let's solve the first equation for i. From the first equation:i = 80 - cThen substitute this into the second equation:5c - 3(80 - c) = 8Let me compute that step by step. First, expand the terms:5c - 3*80 + 3c = 8Wait, wait, no. Wait, hold on. If I distribute the -3 over (80 - c), it would be:5c - 3*80 + 3c = 8Wait, that is:5c - 240 + 3c = 8Wait, combining like terms:5c + 3c is 8c, so:8c - 240 = 8Then, add 240 to both sides:8c = 8 + 2408c = 248Then, divide both sides by 8:c = 248 / 8Let me compute that. 248 divided by 8. 8*30 is 240, so 248 - 240 is 8, so that's 30 + (8/8) = 31. So c is 31.Wait, so the number of correct answers is 31. Let me check if that's right. Let me verify by plugging back into the equations.First, if c is 31, then i = 80 - 31 = 49. Then total points would be 5*31 - 3*49. Let's compute that.5*31: 30*5=150, plus 1*5=5, so 155.3*49: 40*3=120, minus 1*3=3, so 120 - 3 = 117. Wait, wait, no. 49 is 50 -1, so 3*50=150, minus 3*1=3, so 150 -3=147. Wait, wait, hold on, that's conflicting. Let me do 49*3 properly.49*3: 40*3=120, 9*3=27. 120+27=147. Right. So total points: 155 -147 = 8. Which matches the given total. So that seems correct.So, the answer is 31. Hmm. So the student solved 31 problems correctly.But let me just make sure I didn't make any arithmetic mistakes. Let's go through the steps again.Starting with c + i =80 and 5c -3i=8.Express i in terms of c: i=80 -c.Substitute into second equation: 5c -3*(80 - c)=8.Expand: 5c -240 +3c =8. Wait, 5c +3c is 8c, so 8c -240=8.Then 8c=248, so c=248/8=31. Correct. Then i=80-31=49. Then 5*31=155, 3*49=147. 155-147=8. Perfect. So all the steps check out.I think that's solid. So, the answer is 31.Final AnswerThe student solved boxed{31} problems correctly.</think>Let ( c ) be the number of problems solved correctly and ( i ) be the number of problems not solved or solved incorrectly. We know the following:1. The total number of problems is 80: [ c + i = 80 ]2. The total points awarded is 8, with 5 points for each correct answer and -3 points for each incorrect answer: [ 5c - 3i = 8 ]First, solve the first equation for ( i ):[i = 80 - c]Substitute ( i ) into the second equation:[5c - 3(80 - c) = 8]Expand and simplify:[5c - 240 + 3c = 8][8c - 240 = 8][8c = 248][c = frac{248}{8} = 31]Verify the solution by substituting ( c = 31 ) back into the equations:- ( i = 80 - 31 = 49 )- Total points: ( 5 times 31 - 3 times 49 = 155 - 147 = 8 )Both equations are satisfied, confirming the solution is correct.The student solved (boxed{31}) problems correctly.]

🔑:Let c be the number of problems solved correctly and e be the sum of the number of problems solved incorrectly and problems not solved. Therefore, c+e=80 and 5 c-3 e is the number of points the student scored in the evaluation. In the present case,left{begin{aligned}c+e & =80 5 c-3 e & =8end{aligned}right.Solving the system, we find c=31 and e=49. Therefore, the student solved 31 problems correctly.

🧠:[<think>Okay, let's see. I need to figure out which of these sets can form a Pythagorean triple. The options are A, B, C, D. Hmm, Pythagorean triples are sets of three positive integers that satisfy the equation a² + b² = c², right? Where c is the hypotenuse. But wait, some of these options have decimals or fractions, not integers. So maybe the question is a bit broader, asking if they can form a right triangle, even if they aren't integers? Hmm. Let me check each one.Starting with option A: 1, 1, √2. So, let's test if 1² + 1² equals (√2)². Calculating that: 1 + 1 = 2, and (√2)² is also 2. So 2 = 2. That works. But wait, but Pythagorean triples usually refer to integers. So even though they satisfy the equation, are they considered a Pythagorean triple if they aren't integers? The problem says "Pythagorean triple," which traditionally requires all three numbers to be positive integers. So even though these numbers satisfy the Pythagorean theorem, since they aren't all integers, maybe A isn't a valid answer. Hmm, that's confusing. The question is a bit ambiguous. Let me check the other options.Option B: 0.3, 0.4, 0.5. Let's compute 0.3² + 0.4² and see if it's 0.5². 0.09 + 0.16 = 0.25. And 0.5² is 0.25. So that works. Again, similar to A, they satisfy the equation, but are they considered a triple? Since they are decimals, not integers. Maybe the question is just asking if they can form a right triangle regardless of being integers? But the term "Pythagorean triple" typically refers to integers. Let me check the definition again. According to what I remember, a Pythagorean triple consists of three positive integers a, b, c such that a² + b² = c². So non-integers wouldn't qualify. So A and B would not be triples, even though they form right triangles. So maybe the answer is C.Option C: 6, 8, 10. Let's check. 6² + 8² = 36 + 64 = 100. And 10² is 100. So yes, that works. And all numbers are integers. So that's a classic Pythagorean triple. Actually, 6-8-10 is just a multiple of 3-4-5, which is a well-known triple. So C is definitely a Pythagorean triple.Option D: 1/3, 1/4, 1/5. Let's see. Let's check if (1/3)² + (1/4)² equals (1/5)². Calculating: 1/9 + 1/16 = (16 + 9)/144 = 25/144. And (1/5)² is 1/25. 25/144 is approximately 0.1736, and 1/25 is 0.04. Not equal. So that doesn't work. So D is not a Pythagorean triple. Even if we thought of them forming a right triangle, they don't satisfy the equation. So D is out.So the only one that is a Pythagorean triple in the traditional sense is option C. Options A and B form right triangles but aren't triples because they aren't integers, and D doesn't form a right triangle at all. Therefore, the answer should be C.Final Answerboxed{C}</think>To determine which set of numbers can form a Pythagorean triple, we need to check if they satisfy the Pythagorean theorem (a^2 + b^2 = c^2) and if they consist of positive integers.- Option A: (1, 1, sqrt{2}) - Check: (1^2 + 1^2 = (sqrt{2})^2) - Result: (1 + 1 = 2) and ((sqrt{2})^2 = 2). They satisfy the equation but are not all integers. - Conclusion: Not a Pythagorean triple.- Option B: (0.3, 0.4, 0.5) - Check: (0.3^2 + 0.4^2 = 0.5^2) - Result: (0.09 + 0.16 = 0.25) and (0.5^2 = 0.25). They satisfy the equation but are not integers. - Conclusion: Not a Pythagorean triple.- Option C: (6, 8, 10) - Check: (6^2 + 8^2 = 10^2) - Result: (36 + 64 = 100) and (10^2 = 100). They satisfy the equation and are all integers. - Conclusion: A Pythagorean triple.- Option D: (frac{1}{3}, frac{1}{4}, frac{1}{5}) - Check: (left(frac{1}{3}right)^2 + left(frac{1}{4}right)^2 = left(frac{1}{5}right)^2) - Result: (frac{1}{9} + frac{1}{16} = frac{25}{144}) and (left(frac{1}{5}right)^2 = frac{1}{25}). They do not satisfy the equation. - Conclusion: Not a Pythagorean triple.The only set that forms a Pythagorean triple is (boxed{C}).,<think>Okay, so I need to figure out which of these sets can form a Pythagorean triple. Let me recall what a Pythagorean triple is. If I remember correctly, a Pythagorean triple consists of three positive integers a, b, and c, such that a² + b² = c². The key here is that they have to be positive integers, right? So even if some numbers satisfy the equation a² + b² = c², if they aren't all integers, then it's not considered a Pythagorean triple. Let me check each option one by one.Starting with option A: 1, 1, √2. Let me test if 1² + 1² equals (√2)². Calculating that, 1 + 1 = 2, and (√2)² is also 2. So yes, the equation holds. However, the problem is that √2 is not an integer. It's an irrational number. So even though they satisfy the Pythagorean theorem, they aren't all integers. Therefore, set A is not a Pythagorean triple.Moving on to option B: 0.3, 0.4, 0.5. Let me see if 0.3² + 0.4² equals 0.5². Calculating each term: 0.09 + 0.16 = 0.25. And 0.5² is 0.25, so that works. But again, these are decimals, not integers. Even if I convert them to fractions, 0.3 is 3/10, 0.4 is 2/5, and 0.5 is 1/2. Still not integers. So set B also doesn't qualify as a Pythagorean triple because they aren't integers.Next is option C: 6, 8, 10. Let's check if 6² + 8² equals 10². 36 + 64 = 100. 10² is 100. So yes, that works. Now, all three numbers here are integers. So 6, 8, 10 is a valid Pythagorean triple. I think this might be the answer, but let me check the last option just to be sure.Option D: 1/3, 1/4, 1/5. Let's test (1/3)² + (1/4)² and see if it equals (1/5)². Calculating each term: (1/9) + (1/16) = (16 + 9)/144 = 25/144. (1/5)² is 1/25. 25/144 is approximately 0.1736, and 1/25 is 0.04. These are not equal, so the equation doesn't hold. Therefore, set D doesn't satisfy the Pythagorean theorem. Additionally, even if they did, they are fractions, not integers, so it wouldn't be a Pythagorean triple.So, summarizing:- A: Satisfies the equation but not all integers.- B: Satisfies the equation but not all integers.- C: Satisfies the equation and all integers.- D: Doesn't satisfy the equation and not all integers.Therefore, the answer should be option C.Final Answerboxed{C}</think>To determine which set of numbers can form a Pythagorean triple, we need to check if the numbers satisfy the Pythagorean theorem (a^2 + b^2 = c^2) and if all numbers are positive integers.Option A: (1, 1, sqrt{2})- Check: (1^2 + 1^2 = (sqrt{2})^2)- Result: (1 + 1 = 2) (True)- However, (sqrt{2}) is not an integer. Not a Pythagorean triple.Option B: (0.3, 0.4, 0.5)- Check: (0.3^2 + 0.4^2 = 0.5^2)- Result: (0.09 + 0.16 = 0.25) (True)- However, these are decimals, not integers. Not a Pythagorean triple.Option C: (6, 8, 10)- Check: (6^2 + 8^2 = 10^2)- Result: (36 + 64 = 100) (True)- All numbers are integers. This is a Pythagorean triple.Option D: (frac{1}{3}, frac{1}{4}, frac{1}{5})- Check: (left(frac{1}{3}right)^2 + left(frac{1}{4}right)^2 = left(frac{1}{5}right)^2)- Result: (frac{1}{9} + frac{1}{16} = frac{25}{144}) vs. (frac{1}{25} = 0.04)- Not equal. Does not satisfy the theorem. Not a Pythagorean triple.Thus, the correct answer is (boxed{C}).]

🔑:To determine which set of numbers can form a Pythagorean triple, we need to check if the square of the largest number is equal to the sum of the squares of the other two numbers in each set. A Pythagorean triple consists of three positive integers a, b, and c, such that a^2 + b^2 = c^2.Set A: 1, 1, sqrt{2}Checking if it forms a Pythagorean triple:[1^2 + 1^2 stackrel{?}{=} (sqrt{2})^2][1 + 1 = 2]Since sqrt{2} is not a positive integer, this set does not strictly meet the traditional definition of a Pythagorean triple, which requires all members to be positive integers. However, mathematically, it satisfies the condition a^2 + b^2 = c^2.Set B: 0.3, 0.4, 0.5Checking if it forms a Pythagorean triple:[(0.3)^2 + (0.4)^2 stackrel{?}{=} (0.5)^2][0.09 + 0.16 = 0.25]This set satisfies the condition a^2 + b^2 = c^2, but since they are not positive integers, it does not meet the traditional definition of a Pythagorean triple.Set C: 6, 8, 10Checking if it forms a Pythagorean triple:[6^2 + 8^2 stackrel{?}{=} 10^2][36 + 64 = 100]This set satisfies the condition a^2 + b^2 = c^2 and all numbers are positive integers, thus it can form a Pythagorean triple.Set D: frac{1}{3}, frac{1}{4}, frac{1}{5}Checking if it forms a Pythagorean triple:[left(frac{1}{3}right)^2 + left(frac{1}{4}right)^2 stackrel{?}{=} left(frac{1}{5}right)^2][frac{1}{9} + frac{1}{16} = frac{1}{25}]This set satisfies the condition a^2 + b^2 = c^2, but since they are not positive integers, it does not meet the traditional definition of a Pythagorean triple.Therefore, among the given options, only set C meets the criteria for forming a Pythagorean triple with all members being positive integers and satisfying the equation a^2 + b^2 = c^2.The correct answer is: boxed{C}.

🧠:[<think>Alright, so I have this equation: (x^{2} - 4|x| + 3 = t). The problem states that this equation has only three real roots, and I need to find the value of (t). Hmm, okay. Let me think about how to approach this.First, since the equation involves an absolute value term (|x|), it might be helpful to consider the cases where (x geq 0) and (x < 0) separately. That way, I can remove the absolute value and rewrite the equation as two different quadratic equations. Let me try that.For (x geq 0): The absolute value of (x) is just (x), so the equation becomes (x^2 - 4x + 3 = t).For (x < 0): The absolute value of (x) is (-x), so the equation becomes (x^2 + 4x + 3 = t).Okay, so now I have two quadratic equations depending on whether (x) is non-negative or negative. But since we are looking for real roots of the original equation, we need to consider the roots of these two quadratics and ensure that they lie in their respective domains (i.e., (x geq 0) for the first equation and (x < 0) for the second). The original equation will have real roots that are solutions to either of these two quadratics in their respective domains. The total number of real roots will be the sum of the real roots from each quadratic that lie within their domain. The problem states that this total is three. So, we need to figure out for which value of (t) does the combination of roots from both quadratics result in exactly three real roots.Let me visualize the graphs of these functions. The original equation (x^{2} - 4|x| + 3 = t) is a V-shaped graph because of the absolute value. When (x geq 0), it's a quadratic opening upwards with vertex at some point, and similarly for (x < 0), it's another quadratic opening upwards. The horizontal line (y = t) intersects this V-shaped graph. We need the number of intersection points to be exactly three.So, three intersection points would mean that one of the quadratics (either for (x geq 0) or (x < 0)) has two real roots in its domain, and the other quadratic has one real root in its domain. Because three is an odd number, and quadratics can have 0, 1, or 2 roots. Alternatively, maybe both quadratics have two roots each, but one of the roots is at the boundary (x = 0), making it count only once? Wait, but (x = 0) is included in (x geq 0), so if (x = 0) is a root, it would be considered in the first case.But actually, let's check if (x = 0) is a possible root. Plugging (x = 0) into the original equation: (0 - 0 + 3 = t), so (t = 3). So, if (t = 3), then (x = 0) is a root. But does that necessarily give three roots?Wait, perhaps I need to analyze each quadratic equation for their roots in their respective domains.Let me first consider the quadratic for (x geq 0): (x^2 - 4x + 3 = t). Let's write this as (x^2 - 4x + (3 - t) = 0). The roots of this quadratic are given by the quadratic formula: (x = [4 pm sqrt{16 - 4(1)(3 - t)}]/2 = [4 pm sqrt{16 - 12 + 4t}]/2 = [4 pm sqrt{4 + 4t}]/2 = [4 pm 2sqrt{1 + t}]/2 = 2 pm sqrt{1 + t}).Similarly, for the quadratic (x^2 + 4x + 3 = t) (for (x < 0)), rearranged as (x^2 + 4x + (3 - t) = 0). The roots are (x = [-4 pm sqrt{16 - 4(1)(3 - t)}]/2 = [-4 pm sqrt{16 - 12 + 4t}]/2 = [-4 pm sqrt{4 + 4t}]/2 = [-4 pm 2sqrt{1 + t}]/2 = -2 pm sqrt{1 + t}).Therefore, the roots for (x geq 0) are (2 + sqrt{1 + t}) and (2 - sqrt{1 + t}), and for (x < 0) are (-2 + sqrt{1 + t}) and (-2 - sqrt{1 + t}).But we need to check whether these roots lie in their respective domains.Starting with the (x geq 0) case:1. (2 + sqrt{1 + t}): Since (sqrt{1 + t}) is non-negative, this root is (2 +) something non-negative, so it's always greater than or equal to 2. Therefore, it's definitely in the domain (x geq 0).2. (2 - sqrt{1 + t}): Here, we need to ensure that (2 - sqrt{1 + t} geq 0), otherwise, this root would be negative and not in the domain. So, the condition is (2 - sqrt{1 + t} geq 0 implies sqrt{1 + t} leq 2 implies 1 + t leq 4 implies t leq 3).Therefore, for (t leq 3), the quadratic for (x geq 0) has two real roots, and for (t > 3), it has only one real root (since the other root would be negative and thus not in the domain).Now, for the (x < 0) case:1. (-2 + sqrt{1 + t}): Here, (sqrt{1 + t}) is non-negative, so (-2 + sqrt{1 + t}) must be less than 0. Therefore, we require (-2 + sqrt{1 + t} < 0 implies sqrt{1 + t} < 2 implies 1 + t < 4 implies t < 3). If (t geq 3), this root would be non-negative and not in the domain (x < 0).2. (-2 - sqrt{1 + t}): Since both (-2) and (-sqrt{1 + t}) are negative, this root is always less than 0. Therefore, this root is always in the domain (x < 0), regardless of (t).Therefore, for the (x < 0) quadratic, when (t < 3), there are two real roots: (-2 + sqrt{1 + t}) (which is less than 0) and (-2 - sqrt{1 + t}) (always less than 0). When (t geq 3), the root (-2 + sqrt{1 + t}) becomes non-negative, so only one root remains: (-2 - sqrt{1 + t}).So summarizing:- For (t < 3): - (x geq 0) quadratic has two roots: (2 + sqrt{1 + t}) and (2 - sqrt{1 + t}). - (x < 0) quadratic has two roots: (-2 + sqrt{1 + t}) and (-2 - sqrt{1 + t}). - Total real roots: 4.- For (t = 3): - (x geq 0) quadratic: (2 + sqrt{4} = 4) and (2 - sqrt{4} = 0). So roots at x = 4 and x = 0. - (x < 0) quadratic: (-2 + sqrt{4} = 0) (which is not in x < 0) and (-2 - sqrt{4} = -4). So only one root at x = -4. - Total real roots: 4 (from x=4, x=0) and x=-4. Wait, but x=0 is considered in x >=0, and x=-4 in x <0. So total roots are x=4, x=0, x=-4. That's three roots. Wait, so when t=3, there are three real roots?Wait, but according to the previous analysis, for t=3:- In the x >=0 quadratic: two roots, 4 and 0 (but 0 is included in x >=0).- In the x <0 quadratic: roots at 0 and -4, but 0 is not in x <0, so only -4.Therefore, total roots: 4, 0, -4. That is three real roots. So when t=3, the equation has three real roots.But according to the earlier summary, at t=3, x >=0 quadratic has two roots (4 and 0) and x <0 quadratic has one root (-4). So total three roots. Then, in the case of t=3, it's three roots. However, the problem states that the equation has "only three real roots". So is t=3 the answer?But wait, let's check for t >3. For example, t=4.For t=4:- x >=0 quadratic: (2 + sqrt{1 +4} = 2 + sqrt(5) ≈ 4.236), and (2 - sqrt(5) ≈ -0.236). So only one root in x >=0: 4.236.- x <0 quadratic: -2 + sqrt(5) ≈ -2 + 2.236 ≈ 0.236 (not in x <0), and -2 - sqrt(5) ≈ -4.236. So only one root: -4.236.Total roots: two roots (4.236 and -4.236). So for t=4, there are two real roots.For t=2:- x >=0 quadratic: roots at 2 + sqrt(3) ≈3.732 and 2 - sqrt(3)≈0.268. Both positive.- x <0 quadratic: roots at -2 + sqrt(3)≈-0.268 (invalid) and -2 - sqrt(3)≈-3.732 (valid). So two roots in x >=0 and one root in x <0. Wait, but -0.268 is negative? Wait, sqrt(3) is approx 1.732, so -2 +1.732≈-0.268. So yes, that's still negative. Wait, but the x <0 quadratic for t=2:The roots are -2 + sqrt(1 + 2) = -2 + sqrt(3) ≈ -0.268 and -2 - sqrt(3)≈-3.732. So both roots are negative? Wait, sqrt(1 + t) is sqrt(3) ≈1.732, so -2 +1.732 is ≈-0.268, which is negative, and -2 -1.732 is ≈-3.732. So both roots are negative. Therefore, for t=2, the x <0 quadratic has two roots. Therefore, total roots: four.Wait, but that contradicts the previous conclusion that for t <3, there are four roots. Wait, but in t=2, which is less than 3, we have four roots. So when t=3, we have three roots, and for t>3, two roots, and for t<3, four roots.Therefore, the equation has three real roots only when t=3.But let me confirm once again with t=3:Original equation becomes x² -4|x| +3 =3 → x² -4|x|=0 → |x|(|x| -4)=0.So |x|=0 →x=0, or |x|=4→x=4 or x=-4. Thus, roots at x=0,4,-4. So three roots. Therefore, yes, t=3 gives exactly three real roots.But the problem says "has only three real roots". So is t=3 the answer?Wait, but let me check t=0. Then the equation is x² -4|x| +3=0. Let's solve that.For x >=0: x² -4x +3=0 →(x-1)(x-3)=0→x=1,3.For x <0: x² +4x +3=0→(x+1)(x+3)=0→x=-1,-3.So four roots: 1,3,-1,-3. So t=0 gives four roots.Similarly, for t=1:x >=0: x² -4x +2=0. Discriminant: 16 -8=8. Roots: [4 ± sqrt(8)]/2= 2 ± sqrt(2). Both positive.x <0: x² +4x +2=0. Roots: [-4 ± sqrt(8)]/2= -2 ± sqrt(2). So sqrt(2)≈1.414, so -2 +1.414≈-0.586 (valid, since x<0), and -2 -1.414≈-3.414. Both roots are negative. So total four roots.Therefore, for t=1, four roots.So the only time when the equation has three real roots is when t=3. Therefore, the answer is t=3.But let me check another value. Let's take t=4. As before, roots are approximately 4.236 and -4.236. So two roots. So t=4 gives two roots.Wait, but what happens if t is slightly less than 3? Let's say t=2.9.Then, sqrt(1 +2.9)=sqrt(3.9)≈1.974.For x >=0 quadratic: roots 2 ±1.974. So 3.974 and 0.026. Both positive.For x <0 quadratic: roots -2 ±1.974. So -0.026 and -3.974. Both negative.Thus, four roots. Therefore, even when approaching t=3 from below, we still have four roots. At t=3, two roots merge at x=0? Wait, no. Wait, when t=3, the x >=0 quadratic has roots at 4 and 0. The root at 0 comes from 2 - sqrt(1 +3)=2 -2=0. The x <0 quadratic would have roots at -2 + sqrt(4)=0 and -2 - sqrt(4)=-4. But x=0 is not in x <0, so only -4. So at t=3, the x=0 root is counted once (from x >=0), and the roots from x=4 and x=-4. Thus, three roots.If t were greater than 3, say t=4, as before, the x >=0 quadratic only has one root (4.236) and the x <0 quadratic also has one root (-4.236). So total two roots.But here's a thought: maybe when t=3, the root at x=0 is actually a double root or something? Let me check.But x=0 is a root only once. Because when t=3, plugging x=0 into the original equation: 0 -0 +3=3, which equals t=3. So x=0 is a single root. The other roots are x=4 and x=-4. So three distinct real roots.But perhaps the function is tangent to the line y=t at x=0, making it a double root? Let me check the derivative at x=0.But since the function involves |x|, the derivative at x=0 may not exist. Let me check the left and right derivatives.For x >0: the function is x² -4x +3. The derivative is 2x -4. At x=0+, the derivative is -4.For x <0: the function is x² +4x +3. The derivative is 2x +4. At x=0-, the derivative is 4.Therefore, the function has a corner at x=0, with different left and right derivatives. So at x=0, the function isn't differentiable, but the point (0,3) is on the graph when t=3. So when t=3, the horizontal line y=3 touches the graph at x=0 and intersects at x=4 and x=-4. Therefore, x=0 is a single root, not a double root. So even though the line y=3 passes through the point (0,3), which is a "corner" point of the function, it's still a single root. Thus, the equation x² -4|x| +3 =3 has three real roots: 0,4,-4.Therefore, the answer is t=3.But let me confirm once again by analyzing the original function.Let me consider the function f(x) = x² -4|x| +3. We need to find t such that f(x) = t has exactly three real roots.The graph of f(x) is a V-shaped parabola. For x ≥0, it's the parabola x² -4x +3, which has its vertex at x=2 (since vertex x-coordinate is -b/(2a) =4/2=2), and the vertex value is f(2)=4 -8 +3= -1.For x <0, it's the parabola x² +4x +3, which has its vertex at x=-2, and the vertex value is f(-2)=4 -8 +3= -1.Therefore, the graph of f(x) is two parabolas opening upwards, each with vertex at (2, -1) and (-2, -1). So the minimum value of f(x) is -1, achieved at x=2 and x=-2.As t increases from -1 upwards, the horizontal line y=t will intersect the graph:- For t < -1: no real roots.- At t = -1: two real roots at x=2 and x=-2.- For -1 < t <3: the line y=t intersects each parabola twice, so total four real roots.- At t=3: the line y=3 intersects the right parabola at x=4 and x=0, and the left parabola at x=-4. Wait, but x=0 is part of the right parabola.Wait, let's think again. When t=3:For x ≥0: solving x² -4x +3 =3 gives x² -4x =0, x(x-4)=0, roots at x=0 and x=4.For x <0: solving x² +4x +3 =3 gives x² +4x =0, x(x+4)=0, roots at x=0 and x=-4. But x=0 is not in x <0, so only x=-4.Therefore, total roots at x=0,4,-4. Three roots.For t >3: For x ≥0: solving x² -4x +3 =t. The equation is x² -4x + (3 -t)=0. The discriminant is 16 -4*(3 -t)=16 -12 +4t=4 +4t. Which is positive for all t >-1. But the roots are x= [4 ± sqrt(4 +4t)]/2=2 ± sqrt(1 + t). Since t >3, sqrt(1 +t) >2. Therefore, 2 - sqrt(1 + t) <0, so only one root in x ≥0: 2 + sqrt(1 + t).Similarly, for x <0: solving x² +4x +3 =t. The roots are x= [-4 ± sqrt(16 -4*(3 -t))]/2= [-4 ± sqrt(4 +4t)]/2= -2 ± sqrt(1 + t). Here, sqrt(1 +t) >2, so -2 + sqrt(1 + t) >0, which is not in x <0. The other root is -2 - sqrt(1 + t) <0. So only one root in x <0.Therefore, for t >3, two roots: one positive and one negative.For t=3, three roots: 0,4,-4.For t= -1, two roots: 2 and -2.For t between -1 and 3, four roots.For t < -1, no roots.Therefore, the only value of t where the equation has exactly three real roots is t=3.Therefore, the answer is boxed{3}.Final Answerboxed{3}</think>Given the equation (x^2 - 4|x| + 3 = t), we need to find the value of (t) such that the equation has only three real roots.First, we consider the cases for (x geq 0) and (x < 0) separately:1. For (x geq 0): [ x^2 - 4x + 3 = t ] The roots are (2 + sqrt{1 + t}) and (2 - sqrt{1 + t}). These roots must be non-negative. For (2 - sqrt{1 + t}) to be non-negative, (t leq 3).2. For (x < 0): [ x^2 + 4x + 3 = t ] The roots are (-2 + sqrt{1 + t}) and (-2 - sqrt{1 + t}). These roots must be negative. For (-2 + sqrt{1 + t}) to be negative, (t < 3).Analyzing the roots:- For (t < 3), both quadratics have two roots each, leading to four real roots.- For (t = 3): - The quadratic for (x geq 0) has roots (4) and (0). - The quadratic for (x < 0) has roots (-4) (since (-2 + sqrt{4} = 0) is not valid for (x < 0)). - This results in three real roots: (0), (4), and (-4).Thus, the value of (t) that results in exactly three real roots is (boxed{3}).,<think>Okay, so I need to solve this equation: ( x^{2} - 4|x| + 3 = t ), and find the value of ( t ) such that the equation has only three real roots. Hmm, let's think about how to approach this.First, I remember that equations involving absolute values can sometimes be split into cases based on the sign of the variable inside the absolute value. In this case, the absolute value is around ( x ), so maybe I should consider two cases: when ( x geq 0 ) and when ( x < 0 ). But before jumping into that, maybe I can visualize the function ( f(x) = x^{2} - 4|x| + 3 ) and see how it interacts with the horizontal line ( y = t ). The number of real roots will depend on how the horizontal line intersects the graph of ( f(x) ).Let me sketch the graph mentally. The function ( f(x) ) involves ( x^2 ) and ( |x| ), so it's a quadratic in ( x ), but modified by the absolute value. For ( x geq 0 ), ( |x| = x ), so ( f(x) = x^2 - 4x + 3 ). For ( x < 0 ), ( |x| = -x ), so ( f(x) = x^2 - 4(-x) + 3 = x^2 + 4x + 3 ). Therefore, the function is a piecewise combination of two quadratics: one for ( x geq 0 ) and another for ( x < 0 ).Let me write that down:For ( x geq 0 ):( f(x) = x^2 - 4x + 3 )For ( x < 0 ):( f(x) = x^2 + 4x + 3 )Each of these is a parabola. For ( x geq 0 ), the parabola opens upwards (since the coefficient of ( x^2 ) is positive) with vertex at ( x = 2 ). Let me calculate the vertex. The general vertex form for a quadratic ( ax^2 + bx + c ) is at ( x = -b/(2a) ). Here, ( a = 1 ), ( b = -4 ), so vertex at ( x = 4/(2*1) = 2 ). Then, plugging back in, ( f(2) = (2)^2 - 4*(2) + 3 = 4 - 8 + 3 = -1 ). So the vertex is at (2, -1) for the right side.Similarly, for ( x < 0 ), the quadratic is ( x^2 + 4x + 3 ). The vertex here is at ( x = -4/(2*1) = -2 ). Then, ( f(-2) = (-2)^2 + 4*(-2) + 3 = 4 - 8 + 3 = -1 ). So the vertex is at (-2, -1) for the left side.Therefore, the graph of ( f(x) ) is a V-shaped graph composed of two parabolas opening upwards, each with vertices at (2, -1) and (-2, -1). The graph is symmetric with respect to the y-axis because the function involves ( |x| ), which makes it even.Now, the original equation is ( f(x) = t ). So, the number of real roots corresponds to the number of intersections between the horizontal line ( y = t ) and the graph of ( f(x) ). The problem states that there are only three real roots. Let's think about how this can happen.If the horizontal line ( y = t ) cuts the graph at three points, that would mean three intersections. Since the graph is symmetric about the y-axis, usually, except at certain points, the number of roots would be even. For example, if the horizontal line cuts both sides (left and right parabolas), each intersection would correspond to two roots (positive and negative), but if the line is tangent to one of the parabolas or intersects at the vertex, which is on the y-axis (x=0), then we might get an odd number of roots.Wait, but actually, for ( x geq 0 ), the parabola is on the right side, and for ( x < 0 ), it's on the left. At x=0, what is the value of f(x)? Let's compute:For x=0, ( f(0) = 0^2 - 4*|0| + 3 = 0 - 0 + 3 = 3 ). So the point (0,3) is on the graph. However, since the left parabola is for x < 0 and the right for x ≥ 0, at x=0, the right parabola takes over, so the graph is continuous at x=0.Wait, but when x approaches 0 from the left (x < 0), ( f(x) = x^2 + 4x + 3 ). Plugging x=0 here gives 0 + 0 + 3 = 3. Similarly, approaching from the right, x=0, ( f(x) = 0 - 0 + 3 = 3. So, yes, the function is continuous at x=0, and the value is 3.So the graph has two "arms", each being a parabola, opening upwards, each with vertex at (2, -1) and (-2, -1), and meeting at (0, 3). So the graph looks like a "W" that's been squashed in the middle? Wait, no. Each side is a parabola. Let me think. For x ≥ 0, the parabola has vertex at (2, -1), so it starts at x=0, f(0)=3, then goes down to (2, -1), then goes up. Similarly, on the left side, starting at x=0, f(0)=3, going left to x=-2, which is the vertex at (-2, -1), then going up. So the graph is V-shaped in the middle but with parabolas on each side. Wait, actually, maybe it's more like two parabolas glued together at x=0. So the overall shape would have a minimum at x=2 and x=-2, both at y=-1, and the point at x=0 is y=3.So, if we imagine this graph, when we draw the horizontal line y = t, the number of intersections will depend on t.If t is greater than 3, then the line y = t is above the highest point of the graph (which is at x=0, y=3). So, there would be two intersections: one on the right arm (x > 0) and one on the left arm (x < 0). Therefore, two real roots.If t = 3, then the horizontal line touches the graph exactly at x=0. So, there's one root at x=0, and since the line y=3 is tangent to both arms? Wait, let's check:For x ≥ 0, set f(x) = 3: x² -4x + 3 = 3. Then x² -4x = 0 → x(x -4) = 0 → x=0 or x=4. Similarly, for x < 0, f(x) = x² +4x +3 =3 → x² +4x =0 →x(x +4)=0 →x=0 or x=-4. But x=0 is included in both cases, but since x <0 on the left, x=0 is only in the right side. Wait, but when t=3, the equation x² -4|x| +3=3 becomes x² -4|x| =0. Therefore, |x|(x -4 sign(x))=0. So |x|=0 →x=0, or x -4 sign(x)=0. For x >0, sign(x)=1, so x=4. For x <0, sign(x)=-1, so x= -4*(-1)=4? Wait, that might be confusing.Alternatively, solving x² -4|x| =0. Let me factor it as |x|(|x| -4)=0. Because x² = |x|², so |x|² -4|x| =0 → |x|(|x| -4)=0. Therefore, |x|=0 →x=0, or |x|=4 →x=4 or x=-4. Therefore, the roots are x=0, x=4, x=-4. So three roots. Wait, but when t=3, the equation has three real roots. But the problem says "has only three real roots". So is t=3 the answer? Wait, but wait.Wait, let me check again. If t=3, then as above, the roots are x=-4, 0, 4. So three real roots. However, if t is between -1 and 3, the horizontal line y = t would intersect each arm (left and right) twice. For example, take t=0. Then, the equation x² -4|x| +3 =0. Let me solve:For x ≥0: x² -4x +3 =0 → (x-1)(x-3)=0 →x=1 or x=3For x <0: x² +4x +3=0 → (x+1)(x+3)=0 →x=-1 or x=-3Therefore, four roots: -3, -1, 1, 3. So four real roots.If t is between -1 and 3, we get four roots. If t is exactly at -1, then?Set t=-1. Then, the equation x² -4|x| +3 = -1 → x² -4|x| +4 =0.For x ≥0: x² -4x +4=0 → (x-2)^2=0 →x=2For x <0: x² +4x +4=0 → (x+2)^2=0 →x=-2Therefore, two real roots: x=2 and x=-2. But wait, at t=-1, the horizontal line is tangent to both vertices (at x=2 and x=-2). So, each tangent gives a repeated root? So, multiplicity two for each? But in reality, each quadratic is tangent at one point, so two roots each, but each with multiplicity one? Wait, no. For x ≥0, (x-2)^2=0 has root x=2 with multiplicity two, but since x=2 is in x ≥0, it's a single root. Similarly, x=-2 is a root with multiplicity two for x <0. But since we split into cases, each case gives one root. So total two roots. Hmm, so t=-1 gives two real roots.Wait, so when t is less than -1, the equation would have no real roots, because the minimum value of f(x) is -1. So for t < -1, no solutions.For t = -1, two solutions (x=2 and x=-2).For -1 < t < 3, four solutions (two on each side).For t =3, three solutions (x=0, x=4, x=-4).Wait, but when t=3, x=0 is a solution? Let's confirm. Plug x=0 into the equation: 0^2 -4*|0| +3 =0 -0 +3=3. So yes, 3 = t. So x=0 is a solution when t=3, but also x=4 and x=-4. So three solutions. Therefore, when t=3, the equation has three real roots.But the problem states that the equation has "only three real roots". So does that mean t must be 3? But let me check what happens when t >3. For t >3, let's pick t=4. Then the equation x² -4|x| +3 =4 →x² -4|x| -1=0.For x ≥0: x² -4x -1=0. The discriminant is 16 +4=20. So roots x=(4 ±√20)/2=2 ±√5. Since x ≥0, both roots are positive (2 +√5 ≈4.236 and 2 -√5≈-0.236, but negative, so only x≈4.236). Similarly, for x <0: x² +4x -1=0. Discriminant 16 +4=20. Roots x=(-4 ±√20)/2=-2 ±√5. Since x <0, the roots are -2 +√5≈-2+2.236≈0.236 (which is positive, so discard) and -2 -√5≈-4.236. So two roots: x≈4.236 and x≈-4.236. Therefore, two real roots. So for t>3, two real roots. For t=3, three real roots. For -1 <t <3, four real roots. For t=-1, two real roots. For t < -1, no real roots.Therefore, the only value of t where the equation has exactly three real roots is t=3. Therefore, the answer is boxed{3}. Wait, but let me verify again.Wait, when t=3, the roots are x=0, x=4, x=-4. So three roots. However, when t=3, the horizontal line y=3 passes through the point (0,3) and intersects the right and left arms at x=4 and x=-4. But x=4 is on the right arm (x≥0) and x=-4 is on the left arm (x<0). So those are three distinct roots. However, when t=3, is x=0 a root? Yes, because plugging x=0 into the equation gives 0 -0 +3=3=t. So x=0 is a solution, and x=4 and x=-4 as well. So three solutions. For t slightly above 3, say t=3.1, then the horizontal line would intersect each arm once (on the right and left), but miss the middle point at x=0. So two solutions. For t slightly below 3, say t=2.9, the horizontal line would intersect each arm twice, giving four solutions. Therefore, exactly at t=3, the horizontal line is tangent to the middle point (x=0) and intersects the arms at two other points. Wait, but x=0 isn't a tangent point. Wait, the function at x=0 is 3, which is a regular point, not a vertex. The vertices are at (2, -1) and (-2, -1). So when t=3, the line y=3 cuts through the graph at x=0, x=4, and x=-4. So three points. When t is just below 3, the line would intersect the left and right arms each at two points, and pass above x=0, so four roots. When t=3, the line passes through x=0, which is a single root, and cuts each arm at one point (x=4 and x=-4), making three roots total. When t>3, the line is above x=0, so only intersects the arms at two points, x>4 and x<-4. So two roots.Therefore, the only t that gives exactly three real roots is t=3. Therefore, the answer should be 3. So I think that's it.But let me double-check by solving the equation for x.Given ( x^{2} -4|x| +3 = t ). Let me set ( |x| = y ), where ( y geq 0 ). Then the equation becomes ( y^{2} -4y +3 = t ), with ( y = |x| geq 0 ). So solving for y: ( y^2 -4y + (3 - t) =0 ). The quadratic in y is ( y^2 -4y + (3 - t) =0 ). The discriminant is ( 16 -4*(3 - t) =16 -12 +4t=4 +4t=4(t +1) ). So the roots are ( y = [4 ± sqrt(4(t +1))]/2 = [4 ± 2sqrt(t +1)]/2=2 ± sqrt(t +1) ).Therefore, the solutions for y are ( y = 2 + sqrt(t +1) ) and ( y = 2 - sqrt(t +1) ). Since ( y geq 0 ), we need both solutions to be non-negative.First, sqrt(t +1) is real only if t +1 ≥0 → t ≥ -1. So for t < -1, no real solutions. For t ≥ -1, sqrt(t +1) is real.Now, the solutions for y:1. ( y_1 = 2 + sqrt(t +1) ). Since sqrt(t +1) ≥0, y1 ≥2, which is always positive. So this is always a valid solution.2. ( y_2 = 2 - sqrt(t +1) ). We need ( y_2 ≥0 ), so 2 - sqrt(t +1) ≥0 → sqrt(t +1) ≤2 → t +1 ≤4 →t ≤3.Therefore, for t ≤3, y2 is non-negative, so two solutions for y (y1 and y2). For t >3, only y1 is valid.Therefore, for t ≥ -1 and t ≤3, we have two solutions for y: y1 and y2. Then, each y corresponds to two solutions for x (x=±y), except when y=0.Wait, but y is |x|, so for each positive y, there are two real x: ±y. For y=0, there is one real x: x=0.So, let's analyze the number of real roots based on t.Case 1: t < -1: No real roots.Case 2: t = -1: Then sqrt(t +1) = sqrt(0) =0. So y1=2 +0=2, y2=2 -0=2. Wait, that's not right. Wait, if t=-1, then sqrt(t +1)=sqrt(0)=0. Then y1=2 +0=2 and y2=2 -0=2. So both y1 and y2 equal 2. So the quadratic equation in y has a repeated root at y=2. So y=2. Then x=±2. Therefore, two real roots: x=2 and x=-2. So t=-1 gives two real roots.Case 3: -1 < t <3: Then sqrt(t +1) is positive but less than 2 (since t +1 <4). Therefore, y1=2 + sqrt(t +1) and y2=2 - sqrt(t +1). Both y1 and y2 are positive (since sqrt(t +1) <2 in this case, so y2=2 - sqrt(t +1) >0). Therefore, each y gives two real x: ±y1 and ±y2. Therefore, total four real roots.Case 4: t=3: sqrt(t +1)=sqrt(4)=2. Therefore, y1=2 +2=4 and y2=2 -2=0. So solutions y=4 and y=0. For y=4, x=±4. For y=0, x=0. Therefore, three real roots: x=4, x=-4, x=0.Case 5: t >3: sqrt(t +1) >2. Therefore, y1=2 + sqrt(t +1) >2 +2=4, and y2=2 - sqrt(t +1) <0. But y2 must be non-negative, so y2 is invalid. Therefore, only y1 is valid, leading to two real roots: x=±y1=±(2 + sqrt(t +1)).Therefore, summarizing:- t < -1: 0 roots- t = -1: 2 roots (x=±2)- -1 < t <3: 4 roots (x=±(2 + sqrt(t +1)), x=±(2 - sqrt(t +1)))- t =3: 3 roots (x=0, x=±4)- t >3: 2 roots (x=±(2 + sqrt(t +1)))Therefore, the equation has only three real roots only when t=3. Therefore, the answer is boxed{3}.Final Answerboxed{3}</think>Given the equation ( x^2 - 4|x| + 3 = t ), we need to find the value of ( t ) such that the equation has only three real roots.First, we consider the function ( f(x) = x^2 - 4|x| + 3 ). This function can be split into two cases based on the sign of ( x ):- For ( x geq 0 ), ( f(x) = x^2 - 4x + 3 )- For ( x < 0 ), ( f(x) = x^2 + 4x + 3 )Both of these are quadratic functions. The vertex of the quadratic for ( x geq 0 ) is at ( (2, -1) ), and the vertex for ( x < 0 ) is at ( (-2, -1) ). The value at ( x = 0 ) is 3.To analyze the number of real roots of the equation ( f(x) = t ), we consider the horizontal line ( y = t ) intersecting the graph of ( f(x) ). The key points to consider are the vertices at ( (-2, -1) ) and ( (2, -1) ), and the value at ( x = 0 ) which is 3.- For ( t < -1 ), there are no real roots.- For ( t = -1 ), the equation has two real roots.- For ( -1 < t < 3 ), the equation has four real roots.- For ( t = 3 ), the equation has three real roots (x=0, x=4, x=-4).- For ( t > 3 ), the equation has two real roots.Thus, the only value of ( t ) for which the equation has exactly three real roots is ( t = 3 ).[boxed{3}]]

🔑:2, 1.3.The original equation is equivalent to (|x|-2)^{2}=t+1.When t-1, the equation becomes|x|=2+sqrt{t+1} text {, }or |x|=2-sqrt{t+1}.Clearly, equation (1) always has two real roots, so equation (2) must be2-sqrt{t+1}=0 Rightarrow t=3 .

🧠:[<think>Okay, so I need to find the sum of all integer values of x where the function y doesn't exceed 7. The function given is:y = x² + x*(log₅2 - log₂10) - log₂25 - 3 log₅2Hmm, let's start by simplifying the logarithmic terms. Maybe if I can express all logs with the same base or simplify them, it might make the equation easier to handle.First, let's look at log₅2. I know that log_b a can be written as ln a / ln b, but maybe changing base to 2 or 10 might help here. Let's see:log₅2 is equal to (log₂2)/(log₂5) = 1/(log₂5). Similarly, log₂10 is log₂(2*5) = log₂2 + log₂5 = 1 + log₂5. So, log₅2 - log₂10 would be 1/(log₂5) - (1 + log₂5).Let me denote log₂5 as t. Then log₅2 is 1/t, and log₂10 is 1 + t. So, the coefficient of x becomes (1/t - 1 - t). Let's compute that:1/t - t - 1 = (-t² - t + 1)/t. Not sure if that helps yet. Maybe I should compute the numerical values of these logs to make the equation more concrete.Wait, log₂5: since 2^2 = 4 and 2^3 = 8, so log₂5 is approximately 2.3219. Similarly, log₅2 is 1 divided by log₂5, so approximately 0.4307.So, log₅2 - log₂10 ≈ 0.4307 - (log₂10). Let's compute log₂10: log₂(10) ≈ 3.3219. So, 0.4307 - 3.3219 ≈ -2.8912. So, the coefficient of x is approximately -2.8912. Let's note that.Similarly, log₂25: 25 is 5², so log₂25 = log₂5² = 2 log₂5 ≈ 2*2.3219 ≈ 4.6438.Then, 3 log₅2 is 3*(1/log₂5) ≈ 3*0.4307 ≈ 1.2921.Therefore, substituting these approximate values into the equation:y ≈ x² + x*(-2.8912) - 4.6438 - 1.2921 ≈ x² -2.8912x -5.9359So, the function is approximately y ≈ x² - 2.8912x -5.9359. Then we need to find all integers x where this quadratic is ≤7.Alternatively, maybe instead of approximating, I can handle the logarithms symbolically. Let's try that.First, let's re-express all terms in terms of log₂5. Let me set t = log₂5. Then, log₅2 = 1/t, and log₂10 = 1 + t. Also, log₂25 = 2 log₂5 = 2t, and 3 log₅2 = 3/t.So substituting these into the original equation:y = x² + x*(1/t - (1 + t)) - 2t - 3/tSimplify the coefficient of x:1/t -1 - t = -t -1 + 1/tSo, y = x² - (t + 1 - 1/t)x - 2t - 3/tHmm, not sure if that's helpful. Maybe combining terms:Alternatively, perhaps convert all log terms to base 2. Let's try that.Given log₅2 = 1 / log₂5. So, the original equation:y = x² + x*(1/(log₂5) - log₂10) - log₂25 - 3/(log₂5)But log₂10 is log₂(2*5) = 1 + log₂5, so substituting:y = x² + x*(1/(log₂5) - 1 - log₂5) - 2 log₂5 - 3/(log₂5)Let me denote t = log₂5. Then the equation becomes:y = x² + x*(1/t -1 - t) - 2t - 3/tSo, this seems the same as before. Let me write that as:y = x² - x*(t +1 -1/t) -2t -3/tAlternatively, maybe factor out terms with t. Not sure. Maybe the quadratic in x can be expressed as:y = x² - x*(t +1 -1/t) - (2t + 3/t)But perhaps there's a relationship between t and 1/t? Let me compute t + 1/t. Since t = log₂5, which is approximately 2.3219, so t +1/t ≈ 2.3219 + 0.4307 ≈ 2.7526. Not sure if that helps.Alternatively, maybe I can set t = log₂5 and note that 2t is log₂25, and 3/t is 3 log₅2. But perhaps not. Alternatively, maybe consider solving the inequality y ≤7 symbolically.So, starting from:y = x² + x*(log₅2 - log₂10) - log₂25 - 3 log₅2 ≤7Let's rearrange this inequality:x² + x*(log₅2 - log₂10) - log₂25 - 3 log₅2 -7 ≤0So, this is a quadratic inequality in x. To find integer solutions, we need to find the range of x where the quadratic is ≤0, then sum all integers in that interval.Therefore, first, let's find the roots of the quadratic equation:x² + x*(log₅2 - log₂10) - log₂25 - 3 log₅2 -7 =0Let me denote A = log₅2 - log₂10, B = - log₂25 - 3 log₅2 -7So, quadratic equation: x² + A x + B =0We can use the quadratic formula to find the roots:x = [-A ± sqrt(A² -4*1*B)] /2Then, the quadratic is ≤0 between the roots. Therefore, the integer solutions x are the integers between the two roots. Then, we need to compute the roots, find the interval, and sum all integers in that interval.Therefore, let's compute A and B.First, compute A = log₅2 - log₂10As before, log₅2 = 1/(log₂5) = 1/t, and log₂10 = 1 + log₂5 = 1 + t. So, A = 1/t -1 -tSimilarly, B = - log₂25 - 3 log₅2 -7 = -2t -3/t -7So, substituting t = log₂5.Therefore, A = (1 - t - t²)/t, and B = -2t -3/t -7Now, the discriminant D = A² -4*1*BCompute D:D = [ (1 - t - t²)/t ]² -4*( -2t -3/t -7 )Let's compute this step by step.First, expand A²:(1 - t - t²)² / t² = [1 + t² + t⁴ -2t -2t³ + 2t²] / t²? Wait, wait, let's actually compute (1 - t - t²)²:(1 - t - t²)² = (1)^2 + (-t)^2 + (-t²)^2 + 2*(1*(-t) + 1*(-t²) + (-t)*(-t²))= 1 + t² + t⁴ + 2*(-t -t² + t³)= 1 + t² + t⁴ -2t -2t² + 2t³Combine like terms:t⁴ + 2t³ + (t² -2t²) + (-2t) +1= t⁴ + 2t³ - t² -2t +1Therefore, A² = (t⁴ + 2t³ - t² -2t +1)/t²Now, the other part of D is -4B:-4B = -4*(-2t -3/t -7) = 8t +12/t +28Therefore, D = (t⁴ + 2t³ - t² -2t +1)/t² +8t +12/t +28Let's express all terms over t² to combine:= (t⁴ + 2t³ - t² -2t +1)/t² + (8t*t²)/t² + (12/t)*(t)/t +28*t²/t²Wait, maybe better to write each term:8t = 8t * t²/t² = 8t³/t²12/t = 12/t * t²/t² = 12t/t²28 = 28*t²/t²Therefore,D = [t⁴ + 2t³ - t² -2t +1 +8t³ +12t +28t²]/t²Combine the numerator:t⁴ + (2t³ +8t³) + (-t² +28t²) + (-2t +12t) +1= t⁴ +10t³ +27t² +10t +1Therefore, D = (t⁴ +10t³ +27t² +10t +1)/t²Hmm, that numerator looks like a quadratic in t²? Wait, maybe factorable?Let me check: t⁴ +10t³ +27t² +10t +1Let me try to factor this polynomial. Maybe it factors into (t² + a t +1)(t² +b t +1). Let's see:Multiply out (t² + a t +1)(t² +b t +1) = t⁴ + (a + b)t³ + (ab + 2)t² + (a + b)t +1Comparing with t⁴ +10t³ +27t² +10t +1, we have:a + b =10ab +2=27 => ab=25a + b =10 and ab=25. So, solving for a and b:The roots of the equation x² -10x +25=0, which is (x -5)^2=0. Therefore, a=5, b=5.Therefore, the polynomial factors as (t² +5t +1)^2.Therefore, D = (t² +5t +1)^2 / t²Therefore, sqrt(D) = (t² +5t +1)/tTherefore, the roots are:x = [-A ± sqrt(D)] /2 = [ - ( (1 -t -t²)/t ) ± (t² +5t +1)/t ] /2Let me compute numerator:First, compute -A:-A = - (1 -t -t²)/t = (-1 + t + t²)/tThen, sqrt(D) = (t² +5t +1)/tTherefore, the two roots:x1 = [ (-1 + t + t²)/t + (t² +5t +1)/t ] /2x2 = [ (-1 + t + t²)/t - (t² +5t +1)/t ] /2Compute x1:Numerator: (-1 + t + t²) + (t² +5t +1) = (-1 +1) + (t +5t) + (t² +t²) = 0 +6t +2t² = 2t² +6tTherefore, x1 = (2t² +6t)/t /2 = (2t(t +3))/t /2 = (2(t +3))/2 = t +3Similarly, x2:Numerator: (-1 + t + t²) - (t² +5t +1) = (-1 -1) + (t -5t) + (t² -t²) = -2 -4t +0 = -4t -2Therefore, x2 = (-4t -2)/t /2 = (-2(2t +1))/t /2 = (- (2t +1)/t )So, x2 = -(2t +1)/t = -2 -1/tTherefore, the roots are x1 = t +3 and x2 = -2 -1/tBut t = log₂5, so substituting back:x1 = log₂5 +3x2 = -2 -1/(log₂5) = -2 - log₅2Therefore, the quadratic inequality y ≤7 is satisfied for x between x2 and x1, i.e., between -2 - log₅2 and log₂5 +3.Therefore, the integer values of x must satisfy:-2 - log₅2 ≤ x ≤ log₂5 +3Now, let's compute the numerical values to find the integer x's.First, log₂5 ≈2.3219log₅2 =1/log₂5≈0.4307Therefore, x2≈-2 -0.4307≈-2.4307x1≈2.3219 +3≈5.3219Therefore, the interval is approximately from -2.4307 to5.3219Therefore, the integer values of x in this interval are x=-2, -1, 0,1,2,3,4,5Check:-2.4307 ≤x≤5.3219So, x=-2 is included since -2 is greater than -2.4307.x=5 is included since 5 is less than 5.3219.So integers from -2 to5 inclusive.Thus, the integers are: -2, -1, 0,1,2,3,4,5Now, sum these integers.Sum from -2 to5:Sum = (-2) + (-1) +0 +1 +2 +3 +4 +5Compute step by step:Start with (-2) + (-1) = -3-3 +0 = -3-3 +1 = -2-2 +2 =00 +3=33 +4=77 +5=12Therefore, the sum is12.But let me verify once again to ensure I didn't make a mistake in the interval.x must be greater than or equal to x2≈-2.4307 and less than or equal to x1≈5.3219. Therefore, integers x where x≥-2.4307 (so x≥-2 since -2 is greater than -2.4307) and x≤5.3219 (so x≤5). Therefore, integers from -2 to5 inclusive. Correct.So, sum is (-2)+(-1)+0+1+2+3+4+5= (-2-1)+(0+1+2+3+4+5)= (-3)+(15)=12Therefore, the answer is12.But wait, let me check if x=-2 is indeed included. Let me check the original function at x=-2, x=5, and see if y<=7.But perhaps better to check with exact expressions rather than approximations.We can check if x=-2 and x=5 are included.First, let's compute y at x=-2.Original function:y = x² +x*(log₅2 - log₂10) - log₂25 -3 log₅2Compute each term:x=-2:x² =4x*(log₅2 - log₂10)= (-2)*(log₅2 - log₂10)log₅2 - log₂10 = (1/t - (1 + t)) where t=log₂5≈2.3219, which we approximated earlier as≈-2.8912But let's compute log₅2 - log₂10 exactly:log₅2 - log₂10 = (1/log₂5) - (log₂10) = (1/t) - (1 + t) as before. So, plugging t=log₂5.So, at x=-2:First term:4Second term: (-2)*(1/t -1 -t)= (-2)*( (1 - t -t²)/t )Third term: -log₂25 -3 log₅2= -2t -3*(1/t)So, total y=4 + (-2)*( (1 -t -t²)/t ) -2t -3/tLet me compute this expression step by step.First, let's compute (1 -t -t²)/t:= (1/t -1 -t)Then, multiplying by -2:-2*(1/t -1 -t) = -2/t +2 +2tThen, adding the other terms:4 + (-2/t +2 +2t) -2t -3/t =4 +2 + (-2/t -3/t) +2t -2t=6 + (-5/t) +0=6 -5/tBut t=log₂5≈2.3219, so 1/t≈0.4307Therefore, 6 -5*(0.4307)=6 -2.1535≈3.8465Which is less than7. So, y at x=-2≈3.85<=7. Therefore, x=-2 is included.Similarly, check x=5.y=5² +5*(log₅2 - log₂10) -log₂25 -3 log₅2Compute each term:25 +5*( (1/t -1 -t) ) -2t -3/tAgain, compute 5*(1/t -1 -t)=5*(1/t -1 -t)Which is 5*( (1 -t -t²)/t )So, total y=25 +5*( (1 -t -t²)/t ) -2t -3/tCompute step by step:First, 5*(1 -t -t²)/t=5*(1/t -1 -t)=5/t -5 -5tThen, adding the other terms:25 + (5/t -5 -5t) -2t -3/t=25 -5 + (5/t -3/t) + (-5t -2t)=20 +2/t -7tNow, substitute t=log₂5≈2.3219, 2/t≈0.8614, 7t≈16.2533Therefore, y≈20 +0.8614 -16.2533≈20 -15.3919≈4.6081<=7. Therefore, x=5 is included.Therefore, both endpoints x=-2 and x=5 are included, and the integer values in between are indeed from -2 to5 inclusive. Hence, the sum is12.Final Answerboxed{12}</think>To find the sum of all integer values of ( x ) for which the corresponding values of the function[ y = x^2 + x left( log_{5} 2 - log_{2} 10 right) - log_{2} 25 - 3 log_{5} 2 ]do not exceed 7, we start by simplifying the logarithmic terms. Let ( t = log_{2} 5 ). Then, we have:[log_{5} 2 = frac{1}{t}, quad log_{2} 10 = 1 + t, quad log_{2} 25 = 2t, quad text{and} quad 3 log_{5} 2 = frac{3}{t}]Substituting these into the function, we get:[y = x^2 + x left( frac{1}{t} - (1 + t) right) - 2t - frac{3}{t}]Simplifying the coefficients:[y = x^2 + x left( frac{1 - t - t^2}{t} right) - 2t - frac{3}{t}]We need to solve the inequality ( y leq 7 ):[x^2 + x left( frac{1 - t - t^2}{t} right) - 2t - frac{3}{t} - 7 leq 0]This is a quadratic inequality in ( x ). Solving the quadratic equation ( x^2 + x left( frac{1 - t - t^2}{t} right) - 2t - frac{3}{t} - 7 = 0 ) using the quadratic formula, we find the roots:[x = frac{-( frac{1 - t - t^2}{t} ) pm sqrt{ left( frac{1 - t - t^2}{t} right)^2 - 4 left( -2t - frac{3}{t} - 7 right) }}{2}]After simplifying, the roots are found to be ( x_1 = t + 3 ) and ( x_2 = -2 - frac{1}{t} ). Substituting ( t = log_{2} 5 approx 2.3219 ) and ( frac{1}{t} approx 0.4307 ), we approximate the roots:[x_1 approx 2.3219 + 3 approx 5.3219 quad text{and} quad x_2 approx -2 - 0.4307 approx -2.4307]The integer values of ( x ) within the interval ([-2.4307, 5.3219]) are (-2, -1, 0, 1, 2, 3, 4, 5). Summing these integers:[-2 + (-1) + 0 + 1 + 2 + 3 + 4 + 5 = 12]Thus, the sum of all integer values of ( x ) for which the corresponding values of the function do not exceed 7 is (boxed{12}).,<think>Okay, so I need to find the sum of all integer values of x where the function y doesn't exceed 7. The function is given by:y = x² + x(log₅2 − log₂10) − log₂25 − 3 log₅2First, let me parse the function. It's a quadratic in x. The coefficients involve logarithms with different bases. Maybe I can simplify those logarithmic terms first.Starting with the constant terms:log₂25: 25 is 5², so log₂5² = 2 log₂5. So that term becomes 2 log₂5.Then, 3 log₅2: That's straightforward, 3 times log base 5 of 2.The coefficient of x is (log₅2 − log₂10). Hmm, log₂10 can be written as log₂(2*5) = log₂2 + log₂5 = 1 + log₂5. So log₂10 = 1 + log₂5.So substituting that into the coefficient:log₅2 − (1 + log₂5) = log₅2 − 1 − log₂5Now, maybe express all logs in terms of log base 2 or log base 5? Let's see. Let me recall the change of base formula: log_b a = (ln a)/(ln b) = (log_c a)/(log_c b). So, log₅2 = (log₂2)/(log₂5) = 1/(log₂5). Let's denote log₂5 as t. Then log₅2 = 1/t. So, let's see if substituting t helps.Let me set t = log₂5. Then log₅2 = 1/t.Therefore, the coefficient of x becomes:(1/t) - 1 - tAnd the constant terms:-2 log₂5 - 3 log₅2 = -2t - 3*(1/t)So substituting, the function becomes:y = x² + x*(1/t - t - 1) - 2t - 3/tHmm. That might not be immediately helpful. Alternatively, maybe compute numerical values for these logs to simplify?Alternatively, maybe combine terms. Let's see.First, compute log₅2 and log₂5. Let me recall that log₅2 = 1 / log₂5 (since log_b a = 1 / log_a b). So, if I let t = log₂5, then log₅2 = 1/t. So, the coefficient of x is (1/t - t - 1). The constants are -2t - 3/t.Alternatively, maybe express everything in terms of t, and then see if the quadratic can be written in terms of t? Maybe not. Alternatively, let's compute t.Since t = log₂5. Let me approximate log₂5. Since 2² = 4 and 2³ = 8, log₂5 is approximately 2.3219. But maybe we can keep it symbolic for exactness?Alternatively, perhaps express all logs in terms of log base 10 or natural logs? Wait, maybe there is a relationship between log₂5 and log₅2. As I said, log₅2 = 1 / log₂5. So if I let t = log₂5, then log₅2 = 1/t.So, the coefficient of x is (1/t - log₂10). Wait, log₂10 is 1 + log₂5 = 1 + t. So coefficient is (1/t - (1 + t)) = 1/t -1 - t.Similarly, the constants are -2t -3*(1/t) = -2t - 3/t.So, the quadratic is:y = x² + x*(1/t -1 - t) -2t -3/tBut I don't know, maybe this substitution isn't helpful. Let me try to compute numerical values for t, then substitute back.Given that t = log₂5 ≈ 2.321928095Therefore, 1/t ≈ 0.4306765581So, the coefficient of x:1/t -1 - t ≈ 0.4306765581 -1 -2.321928095 ≈ 0.4306765581 -3.321928095 ≈ -2.891251537Similarly, the constants:-2t -3/t ≈ -2*(2.321928095) -3*(0.4306765581) ≈ -4.64385619 -1.292029674 ≈ -5.935885864Therefore, the quadratic is approximately:y ≈ x² -2.891251537x -5.935885864But maybe that's messy. Alternatively, maybe keep it symbolic.Alternatively, perhaps notice that log₂25 is 2 log₂5, and 3 log₅2 is 3*(1 / log₂5). So perhaps if we set t = log₂5, then the entire equation can be written in terms of t. Let's see.Original function:y = x² + x*(log₅2 - log₂10) - log₂25 -3 log₅2Expressed as:y = x² + x*(1/t - (1 + t)) - 2t - 3/tSo, combining terms:y = x² + x*( (1/t - t -1) ) -2t -3/tBut I'm not sure if this helps. Alternatively, maybe factor the expression.Alternatively, set y ≤7, so:x² + x*(log₅2 - log₂10) - log₂25 -3 log₅2 ≤7Bring 7 to the left:x² + x*(log₅2 - log₂10) - log₂25 -3 log₅2 -7 ≤0So, this is a quadratic inequality. To find the integer x values where this holds, we can first find the roots of the quadratic equation x² + x*(log₅2 - log₂10) - log₂25 -3 log₅2 -7 =0, then determine the intervals where the quadratic is ≤0, and then sum the integer x's in that interval.But perhaps instead of approximating the logs, which might lead to inaccuracies, try to manipulate the logarithmic terms symbolically.Let me try to compute the coefficients step by step.First, let's compute log₅2 - log₂10:log₅2 = 1 / log₂5 (change of base formula)log₂10 = log₂(2*5) = log₂2 + log₂5 = 1 + log₂5Therefore, log₅2 - log₂10 = (1 / log₂5) - (1 + log₂5)Let me denote a = log₂5. Then, the expression becomes (1/a) -1 -aSimilarly, the constants: -log₂25 -3 log₅2 = -2 log₂5 -3*(1 / log₂5) = -2a -3*(1/a)So, the quadratic inequality is:x² + x*(1/a -1 -a) -2a -3/a -7 ≤0But this seems a bit too abstract. Maybe plug back in terms of a:Let me write the quadratic as:x² + x*( (1 - a -a²)/a ) - (2a + 3/a +7 ) ≤0Not sure. Alternatively, perhaps this quadratic can be written in terms of a common denominator.Alternatively, multiply through by a to eliminate denominators. But since a = log₂5 ≈2.3219 is positive, multiplying the inequality by a will preserve the inequality direction.So, multiplying both sides by a:a x² + x*(1 -a -a²) -2a² -3 -7a ≤0Therefore:a x² + (1 -a -a²)x - (2a² +7a +3) ≤0Hmm. Let's substitute a = log₂5. But unless there's a relation here, perhaps not helpful. Alternatively, maybe factor the quadratic.Wait, perhaps factor the constants:Looking at 2a² +7a +3. Let's check if this factors:2a² +7a +3. Let's see. The discriminant is 49 -24 =25. So roots at (-7 ±5)/4. That is, (-7+5)/4 = -0.5 and (-7-5)/4=-3. So factors are (2a +1)(a +3). Let me check:(2a +1)(a +3) = 2a² +6a +a +3 = 2a² +7a +3. Yes.So 2a² +7a +3 = (2a +1)(a +3). Therefore, the quadratic inequality becomes:a x² + (1 -a -a²)x - (2a +1)(a +3) ≤0But not sure. Maybe the quadratic factors? Let me check the quadratic in x:a x² + (1 -a -a²)x - (2a +1)(a +3)Let me denote coefficients:A = aB = 1 -a -a²C = - (2a +1)(a +3)So discriminant D = B² -4ACCompute D:= (1 -a -a²)^2 -4*a*(- (2a +1)(a +3))First compute (1 -a -a²)^2:Let me expand:(1 -a -a²)^2 = ( -a² -a +1 )^2= (-a²)^2 + (-a)^2 +1^2 + 2*(-a²)*(-a) + 2*(-a²)*1 + 2*(-a)*1= a^4 + a² +1 + 2a³ -2a² -2a= a^4 + 2a³ -a² -2a +1Then compute 4AC:4*a*(- (2a +1)(a +3)) = -4a*(2a +1)(a +3)First compute (2a +1)(a +3) = 2a² +6a +a +3 = 2a² +7a +3Thus, 4AC = -4a*(2a² +7a +3) = -8a³ -28a² -12aTherefore, discriminant D = (a^4 + 2a³ -a² -2a +1) -8a³ -28a² -12aCombine like terms:a^4 +2a³ -a² -2a +1 -8a³ -28a² -12a= a^4 + (2a³ -8a³) + (-a² -28a²) + (-2a -12a) +1= a^4 -6a³ -29a² -14a +1Hmm, this seems complicated. Maybe not useful. Alternatively, perhaps the quadratic factors?Assuming that the quadratic factors as (mx +n)(px +q), but given the coefficients in terms of a, it's hard to see. Maybe this approach is not helpful.Alternatively, perhaps plug the value of a numerically and compute the roots?Since a = log₂5 ≈2.321928095Compute coefficients:A = a ≈2.321928B =1 -a -a² ≈1 -2.321928 - (2.321928)^2 ≈1 -2.321928 -5.3915 ≈1 -7.7134 ≈-6.7134C = - (2a +1)(a +3) = - (2*2.321928 +1)(2.321928 +3) ≈- (4.643856 +1)(5.321928) ≈-5.643856*5.321928 ≈-5.643856*5.321928 ≈-30.0508So quadratic equation is approximately:2.321928 x² -6.7134 x -30.0508 ≤0To solve 2.321928 x² -6.7134 x -30.0508 ≤0First find roots:x = [6.7134 ± sqrt(6.7134² -4*2.321928*(-30.0508))]/(2*2.321928)Compute discriminant:D = 6.7134² -4*2.321928*(-30.0508) ≈45.06 + 4*2.321928*30.0508 ≈45.06 + 4*69.777≈45.06 +279.108≈324.168sqrt(D)≈18.0So roots:x ≈ [6.7134 ±18.0]/4.643856First root: (6.7134 +18)/4.643856≈24.7134/4.643856≈5.32Second root: (6.7134 -18)/4.643856≈-11.2866/4.643856≈-2.43Thus, the quadratic is ≤0 between x ≈-2.43 and x≈5.32Since the quadratic opens upwards (A≈2.32>0), it is below zero between the roots.Therefore, the solution set is x ∈ [-2.43,5.32]But x must be integer. So integer values are from -2 to 5 inclusive. Wait, let's check:-2.43 is between -3 and -2. So the smallest integer greater than or equal to -2.43 is -2.Similarly, 5.32 is between 5 and 6, so the largest integer less than or equal to 5.32 is 5.Thus, integer x from -2 to 5 inclusive. Let's list them: -2, -1, 0, 1, 2, 3, 4, 5.Sum these integers:Sum = (-2) + (-1) +0 +1 +2 +3 +4 +5Calculate step by step:(-2) + (-1) = -3-3 +0 = -3-3 +1 = -2-2 +2 =00 +3=33 +4=77 +5=12So sum is 12.But wait, this is based on approximate roots. But perhaps the exact roots might include or exclude some integers? Let's check with exact expressions.Alternatively, maybe compute the exact roots using symbolic computation.Wait, let's recall that a = log₂5. So perhaps there's an exact expression. But this seems complicated.Alternatively, maybe test the integers around -2.43 and 5.32.So the approximate roots are -2.43 and 5.32.Thus, integers from -2 to 5.But maybe check x=-2 and x=5 to see if they satisfy y ≤7.Compute y at x=-2 and x=5 using original function.First, compute the original function:y =x² +x(log₅2 − log₂10) − log₂25 −3 log₅2First, compute constants:log₅2 = 1 / log₂5 ≈0.4306765581log₂10 =1 + log₂5 ≈1 +2.321928≈3.321928log₂25=2 log₂5≈4.6438563 log₅2≈3*0.430676≈1.292029So, the function becomes:y =x² +x*(0.4306765581 -3.321928) -4.643856 -1.292029Compute coefficients:0.4306765581 -3.321928 ≈-2.89125144-4.643856 -1.292029 ≈-5.935885Thus, y ≈x² -2.89125144x -5.935885Compute y at x=-2:y≈(-2)^2 -2.89125144*(-2) -5.935885≈4 +5.78250288 -5.935885≈4 +5.7825 -5.935885≈(4 -5.935885) +5.7825≈-1.935885 +5.7825≈3.846615Which is approximately 3.85 ≤7, so x=-2 is included.x=5:y≈5² -2.89125144*5 -5.935885≈25 -14.4562572 -5.935885≈25 -20.3921422≈4.6078578≈4.61 ≤7, so x=5 is included.Check x=6:y≈6² -2.89125144*6 -5.935885≈36 -17.3475086 -5.935885≈36 -23.2833936≈12.7166064≈12.72 >7, so excluded.Check x=-3:y≈(-3)^2 -2.89125144*(-3) -5.935885≈9 +8.67375432 -5.935885≈9 +8.673754 -5.935885≈17.673754 -5.935885≈11.737869≈11.74 >7, so excluded.Therefore, indeed integer x from -2 to 5 inclusive.Sum them: (-2)+(-1)+0+1+2+3+4+5= (-2-1)+(0+1+2+3+4+5)= (-3)+15=12.But wait, the problem says "sum of all integer values of the argument x for which the corresponding values of the function... do not exceed 7."But to be thorough, maybe check the exact values for x=-2, -1,0,1,2,3,4,5 using the original function with exact logarithmic expressions, to confirm that y ≤7.Alternatively, use exact expressions. Let's attempt that.Original function:y =x² +x(log₅2 − log₂10) − log₂25 −3 log₅2First, express all logs in terms of log₂5. Let t = log₂5.Thus:log₅2 = 1/tlog₂10 =1 + tlog₂25=2t3 log₅2=3/tTherefore, the function becomes:y =x² +x*(1/t - (1 + t)) -2t -3/t= x² +x*(1/t -1 -t) -2t -3/tSo, y = x² -x(1 + t -1/t) -2t -3/tBut it's complicated. Let's compute y exactly for each integer x from -2 to 5.First, x=-2:y = (-2)^2 + (-2)*(log₅2 - log₂10) - log₂25 -3 log₅2=4 + (-2)*(log₅2 - log₂10) - log₂25 -3 log₅2=4 -2 log₅2 + 2 log₂10 - log₂25 -3 log₅2Combine like terms:=4 +2 log₂10 - log₂25 -5 log₅2Substitute log₂10 =1 + log₂5, log₂25=2 log₂5, log₅2=1/log₂5.=4 +2(1 + log₂5) -2 log₂5 -5*(1/log₂5)Simplify:=4 +2 +2 log₂5 -2 log₂5 -5/log₂5=6 -5/log₂5Since log₂5 ≈2.3219, so 5/log₂5 ≈5/2.3219≈2.153Thus, y≈6 -2.153≈3.847, which is ≤7.Similarly, x=-2 is okay.x=-1:y=(-1)^2 + (-1)*(log₅2 - log₂10) - log₂25 -3 log₅2=1 - (log₅2 - log₂10) - log₂25 -3 log₅2=1 -log₅2 + log₂10 - log₂25 -3 log₅2=1 + log₂10 - log₂25 -4 log₅2Again substitute:=1 + (1 + log₂5) -2 log₂5 -4/log₂5=1 +1 + log₂5 -2 log₂5 -4/log₂5=2 - log₂5 -4/log₂5Compute numerically:log₂5≈2.3219, so 2 -2.3219 -4/2.3219≈2 -2.3219 -1.723≈2 -4.0449≈-2.0449So y≈-2.0449 ≤7. So included.x=0:y=0 +0 - log₂25 -3 log₅2= -2 log₂5 -3/log₂5≈-2*2.3219 -3/2.3219≈-4.6438 -1.292≈-5.9358 ≤7.Included.x=1:y=1 +1*(log₅2 - log₂10) - log₂25 -3 log₅2=1 + log₅2 - log₂10 - log₂25 -3 log₅2=1 - log₂10 - log₂25 -2 log₅2Substitute:=1 - (1 + log₂5) -2 log₂5 -2/log₂5=1 -1 - log₂5 -2 log₂5 -2/log₂5= -3 log₂5 -2/log₂5≈-3*2.3219 -2/2.3219≈-6.9657 -0.861≈-7.8267 ≤7. Included.x=2:y=4 +2*(log₅2 - log₂10) - log₂25 -3 log₅2=4 +2 log₅2 -2 log₂10 - log₂25 -3 log₅2=4 -2 log₂10 - log₂25 - log₅2Substitute:=4 -2(1 + log₂5) -2 log₂5 -1/log₂5=4 -2 -2 log₂5 -2 log₂5 -1/log₂5=2 -4 log₂5 -1/log₂5≈2 -4*2.3219 -1/2.3219≈2 -9.2876 -0.4307≈-7.7183 ≤7. Included.x=3:y=9 +3*(log₅2 - log₂10) - log₂25 -3 log₅2=9 +3 log₅2 -3 log₂10 - log₂25 -3 log₅2=9 -3 log₂10 - log₂25Substitute:=9 -3(1 + log₂5) -2 log₂5=9 -3 -3 log₂5 -2 log₂5=6 -5 log₂5≈6 -5*2.3219≈6 -11.6095≈-5.6095 ≤7. Included.x=4:y=16 +4*(log₅2 - log₂10) - log₂25 -3 log₅2=16 +4 log₅2 -4 log₂10 - log₂25 -3 log₅2=16 -4 log₂10 - log₂25 + log₅2Substitute:=16 -4(1 + log₂5) -2 log₂5 +1/log₂5=16 -4 -4 log₂5 -2 log₂5 +1/log₂5=12 -6 log₂5 +1/log₂5≈12 -6*2.3219 +0.4307≈12 -13.9314 +0.4307≈12 -13.5007≈-1.5007 ≤7. Included.x=5:y=25 +5*(log₅2 - log₂10) - log₂25 -3 log₅2=25 +5 log₅2 -5 log₂10 - log₂25 -3 log₅2=25 -5 log₂10 - log₂25 +2 log₅2Substitute:=25 -5(1 + log₂5) -2 log₂5 +2/log₂5=25 -5 -5 log₂5 -2 log₂5 +2/log₂5=20 -7 log₂5 +2/log₂5≈20 -7*2.3219 +2/2.3219≈20 -16.2533 +0.861≈20 -15.3923≈4.6077 ≤7. Included.So all x from -2 to 5 inclusive give y ≤7. Therefore, sum is 12.Thus, the answer is boxed{12}.Final Answerboxed{12}</think>To find the sum of all integer values of ( x ) for which the function [y = x^2 + xleft(log _{5} 2 - log _{2} 10right) - log _{2} 25 - 3 log _{5} 2]does not exceed 7, we start by simplifying the logarithmic terms. First, we express the logarithmic terms in terms of ( log_2 5 ):- ( log_5 2 = frac{1}{log_2 5} )- ( log_2 10 = 1 + log_2 5 )- ( log_2 25 = 2 log_2 5 )- ( 3 log_5 2 = frac{3}{log_2 5} )Let ( t = log_2 5 ). Substituting these into the function, we get:[y = x^2 + xleft(frac{1}{t} - (1 + t)right) - 2t - frac{3}{t}]Simplifying the coefficients:- The coefficient of ( x ) is ( frac{1}{t} - 1 - t )- The constant term is ( -2t - frac{3}{t} )We approximate ( t approx 2.321928 ) and substitute back to get the quadratic inequality:[x^2 - 2.891251537x - 5.935885864 leq 7]Solving the quadratic inequality ( x^2 - 2.891251537x - 5.935885864 leq 7 ):- Subtract 7 from both sides: ( x^2 - 2.891251537x - 12.935885864 leq 0 )- Find the roots using the quadratic formula: approximately ( x approx -2.43 ) and ( x approx 5.32 )The integer values of ( x ) within the interval ([-2.43, 5.32]) are (-2, -1, 0, 1, 2, 3, 4, 5). Summing these integers:[(-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 = 12]Thus, the sum of all integer values of ( x ) for which the function does not exceed 7 is (boxed{12}).]

🔑:Solution. Let a=log _{2} 5. Then the condition of the problem will turn into the inequalityx^{2}-left(a-frac{1}{a}+1right) x-left(2 a+frac{3}{a}+7right) leqslant 0 .Considering that a in(2,3), we get x inleft[-frac{1}{a}-2, a+3right]. Since -3<-frac{1}{a}-2<-2,5<a+3<6, the integer solutions will be the numbers -2,-1,0,1,2,3,4,5.Answer: 12. (D)

🧠:[<think>Okay, so I need to prove that if a is positive, then the expression sqrt(a² + 1/a²) + 2 is greater than or equal to a + 1/a + sqrt(2). Hmm, let me start by understanding what each part of this inequality represents.First, let's denote x = a + 1/a. Since a > 0, by AM-GM inequality, a + 1/a is at least 2. Because the arithmetic mean of a and 1/a is (a + 1/a)/2, which is greater than or equal to the geometric mean sqrt(a * 1/a) = 1. So, multiplying both sides by 2 gives a + 1/a ≥ 2. So, x ≥ 2.Now, the left-hand side (LHS) of the inequality is sqrt(a² + 1/a²) + 2. Let me see if I can express a² + 1/a² in terms of x. Since x = a + 1/a, squaring both sides gives x² = a² + 2*(a*(1/a)) + 1/a² = a² + 2 + 1/a². Therefore, a² + 1/a² = x² - 2. So, the LHS becomes sqrt(x² - 2) + 2.The right-hand side (RHS) of the inequality is a + 1/a + sqrt(2) = x + sqrt(2). So, the inequality to prove reduces to:sqrt(x² - 2) + 2 ≥ x + sqrt(2), where x ≥ 2.So now, instead of dealing with a, we can just consider x ≥ 2. Let's check if this transformed inequality holds for all x ≥ 2.Let me denote the left-hand side as L = sqrt(x² - 2) + 2 and the right-hand side as R = x + sqrt(2). We need to show that L ≥ R.Let me rearrange the inequality:sqrt(x² - 2) + 2 - x - sqrt(2) ≥ 0.Let me denote this difference as D = sqrt(x² - 2) + 2 - x - sqrt(2). We need to show that D ≥ 0 for all x ≥ 2.To analyze D, maybe I can consider substituting x = 2 first. Let's check if the inequality holds at x = 2.At x = 2: sqrt(4 - 2) + 2 - 2 - sqrt(2) = sqrt(2) + 2 - 2 - sqrt(2) = 0. So, D = 0 when x = 2. That's a good start.Now, maybe we can check the derivative of D with respect to x and see if D is increasing or decreasing for x ≥ 2.Compute dD/dx:d/dx [sqrt(x² - 2)] = (1/(2*sqrt(x² - 2)))*2x = x / sqrt(x² - 2).Then derivative of 2 is 0, derivative of (-x) is -1, derivative of (-sqrt(2)) is 0. So, total derivative:dD/dx = x / sqrt(x² - 2) - 1.We need to check if this derivative is positive or negative for x > 2.Let me compute x / sqrt(x² - 2) - 1. Let's denote this as D'.If D' ≥ 0, then D is increasing, which would mean that since D(2) = 0, D(x) ≥ 0 for x ≥ 2. If D' ≤ 0, then D is decreasing, which would be a problem because we need D(x) ≥ 0.So, let's see:x / sqrt(x² - 2) - 1. Let's rationalize this expression.Multiply numerator and denominator by sqrt(x² - 2):Wait, perhaps instead, let's set t = x². Then, x / sqrt(x² - 2) = x / sqrt(x² - 2). Let me square the term x / sqrt(x² - 2):(x / sqrt(x² - 2))² = x² / (x² - 2). Since x > 2, x² - 2 > 0, so this is valid.So, x² / (x² - 2). Compare this to 1. So, x² / (x² - 2) = 1 + 2 / (x² - 2). Since x ≥ 2, x² - 2 ≥ 4 - 2 = 2. Therefore, 2/(x² - 2) ≤ 1. Thus, x²/(x² - 2) ≤ 1 + 1 = 2. Wait, but that might not be helpful.Alternatively, perhaps compare x / sqrt(x² - 2) to 1.Let me compute x / sqrt(x² - 2):Let x > 2. Then sqrt(x² - 2) = x * sqrt(1 - 2/x²). Therefore, x / sqrt(x² - 2) = 1 / sqrt(1 - 2/x²). Since 2/x² < 1 for x > sqrt(2), which is certainly true since x ≥ 2. So, sqrt(1 - 2/x²) is less than 1, so 1 / sqrt(1 - 2/x²) is greater than 1. Therefore, x / sqrt(x² - 2) > 1. Therefore, D' = x / sqrt(x² - 2) - 1 > 0.Thus, the derivative D' is positive for x > 2, which means that D(x) is increasing for x ≥ 2. Since D(2) = 0 and D is increasing for x > 2, D(x) ≥ 0 for all x ≥ 2. Therefore, the original inequality holds.But let me verify with another approach to ensure that this is correct.Alternatively, maybe we can square both sides of the inequality to eliminate the square root. However, since we are dealing with inequalities, squaring can sometimes be tricky because we have to ensure both sides are positive. Let's check if that's feasible here.Starting from the transformed inequality:sqrt(x² - 2) + 2 ≥ x + sqrt(2).Since x ≥ 2, sqrt(x² - 2) is real and positive, and x + sqrt(2) is also positive. So, both sides are positive, so squaring should preserve the inequality.Let me square both sides:(sqrt(x² - 2) + 2)² ≥ (x + sqrt(2))².Left side: (sqrt(x² - 2))² + 4*sqrt(x² - 2) + 4 = (x² - 2) + 4*sqrt(x² - 2) + 4 = x² + 2 + 4*sqrt(x² - 2).Right side: x² + 2*sqrt(2)*x + 2.So, the inequality becomes:x² + 2 + 4*sqrt(x² - 2) ≥ x² + 2*sqrt(2)*x + 2.Subtracting x² + 2 from both sides:4*sqrt(x² - 2) ≥ 2*sqrt(2)*x.Divide both sides by 2:2*sqrt(x² - 2) ≥ sqrt(2)*x.Divide both sides by sqrt(2):sqrt(2)*sqrt(x² - 2) ≥ x.Square both sides again to eliminate the square roots:2*(x² - 2) ≥ x².Simplify left side: 2x² - 4 ≥ x².Subtract x² from both sides:x² - 4 ≥ 0.Which simplifies to x² ≥ 4. Since x ≥ 2, x² ≥ 4 is true. Therefore, all the previous steps are valid, and thus the original inequality holds.Therefore, the inequality sqrt(a² + 1/a²) + 2 ≥ a + 1/a + sqrt(2) holds for all a > 0.But let me check with a specific example to ensure that there are no mistakes. Let's take a = 1. Then, LHS: sqrt(1 + 1) + 2 = sqrt(2) + 2 ≈ 1.414 + 2 = 3.414. RHS: 1 + 1 + sqrt(2) ≈ 2 + 1.414 ≈ 3.414. So, equality holds here. But according to our earlier analysis, equality should hold when x = 2, which corresponds to a + 1/a = 2, which happens when a = 1. So, that's consistent.Another test case: Let a = 2. Then, LHS: sqrt(4 + 1/4) + 2 = sqrt(17/4) + 2 = (√17)/2 + 2 ≈ 4.123/2 + 2 ≈ 2.0615 + 2 = 4.0615. RHS: 2 + 1/2 + sqrt(2) ≈ 2.5 + 1.414 ≈ 3.914. So, 4.0615 ≥ 3.914, which is true.Another test case: a approaching infinity. Let a be very large, then 1/a is negligible. Then, LHS: sqrt(a² + 0) + 2 ≈ a + 2. RHS: a + 0 + sqrt(2) ≈ a + 1.414. So, a + 2 ≥ a + 1.414, which is true.Similarly, when a approaches 0 from the positive side, 1/a becomes very large. Let a approach 0, then a is approaching 0, 1/a approaches infinity. Then, LHS: sqrt( (1/a²) + a² ) + 2 ≈ sqrt(1/a²) + 2 = 1/a + 2. RHS: a + 1/a + sqrt(2) ≈ 1/a + sqrt(2). So, 1/a + 2 ≥ 1/a + sqrt(2) → 2 ≥ sqrt(2), which is true. So, the inequality holds in the limit as a approaches 0 and infinity.Therefore, based on both analytical methods and test cases, the inequality seems valid. The key steps were substituting x = a + 1/a, transforming the inequality, then either analyzing the derivative or squaring both sides to reduce it to a simpler inequality. Both approaches led to the conclusion that the inequality holds for all a > 0.Final Answerboxed{sqrt{a^2 + frac{1}{a^2}} + 2 geq a + frac{1}{a} + sqrt{2}}</think>Given ( a > 0 ), we need to prove that:[sqrt{a^2 + frac{1}{a^2}} + 2 geq a + frac{1}{a} + sqrt{2}]First, let ( x = a + frac{1}{a} ). By the AM-GM inequality, ( x geq 2 ).Next, we express ( a^2 + frac{1}{a^2} ) in terms of ( x ):[x^2 = left( a + frac{1}{a} right)^2 = a^2 + 2 + frac{1}{a^2} implies a^2 + frac{1}{a^2} = x^2 - 2]Substituting this into the left-hand side (LHS) of the inequality, we get:[sqrt{x^2 - 2} + 2]The right-hand side (RHS) of the inequality becomes:[x + sqrt{2}]Thus, the inequality transforms to:[sqrt{x^2 - 2} + 2 geq x + sqrt{2}]We need to show this holds for ( x geq 2 ). Let ( D = sqrt{x^2 - 2} + 2 - x - sqrt{2} ). We check if ( D geq 0 ).First, at ( x = 2 ):[sqrt{4 - 2} + 2 - 2 - sqrt{2} = sqrt{2} + 2 - 2 - sqrt{2} = 0]Next, we compute the derivative ( D' ):[D' = frac{x}{sqrt{x^2 - 2}} - 1]Since ( x / sqrt{x^2 - 2} > 1 ) for ( x > 2 ), ( D' > 0 ). Thus, ( D ) is increasing for ( x geq 2 ). Since ( D(2) = 0 ) and ( D ) is increasing, ( D(x) geq 0 ) for ( x geq 2 ).Alternatively, squaring both sides of the inequality:[(sqrt{x^2 - 2} + 2)^2 geq (x + sqrt{2})^2]Expanding both sides:[x^2 + 2 + 4sqrt{x^2 - 2} geq x^2 + 2sqrt{2}x + 2]Simplifying:[4sqrt{x^2 - 2} geq 2sqrt{2}x implies 2sqrt{x^2 - 2} geq sqrt{2}x implies sqrt{2}sqrt{x^2 - 2} geq x]Squaring again:[2(x^2 - 2) geq x^2 implies x^2 geq 4]Since ( x geq 2 ), ( x^2 geq 4 ) holds. Thus, the inequality is proven.Therefore, the final answer is:[boxed{sqrt{a^2 + frac{1}{a^2}} + 2 geq a + frac{1}{a} + sqrt{2}}],<think>Okay, so I need to prove that if a is a positive real number, then the inequality √(a² + 1/a²) + 2 is greater than or equal to a + 1/a + √2. Hmm. Let me think about how to approach this.First, maybe I can simplify the expression on the left side and see if I can relate it to the right side. Let's write down both sides clearly:Left side: √(a² + 1/a²) + 2Right side: a + 1/a + √2I need to show that Left side - Right side ≥ 0. So maybe subtract the right side from the left side and show that the result is non-negative.Let me denote L = √(a² + 1/a²) + 2 and R = a + 1/a + √2. So we need to prove L - R ≥ 0.Alternatively, perhaps square both sides? But since square is a monotonic function for non-negative numbers, maybe that would work, but I need to be careful because squaring can sometimes introduce extraneous solutions.Wait, but both sides are positive. Since a > 0, a + 1/a is at least 2 by AM ≥ GM. The left side has √(a² + 1/a²) which is at least √(2*(a)*(1/a)) = √2, again by AM ≥ GM. So the left side is at least √2 + 2, and the right side is at least 2 + √2. Wait, that's interesting. Both sides have a minimum of 2 + √2? But if that's the case, then maybe equality holds when a = 1/a, which is when a = 1. Let me check:If a = 1, left side is √(1 + 1) + 2 = √2 + 2. Right side is 1 + 1 + √2 = 2 + √2. So equality holds here. So maybe the inequality is tight at a = 1. Now, we need to show that for all a > 0, the left side is greater than or equal to the right side.Alternatively, maybe substitute t = a + 1/a. Let me try that. Since a > 0, t ≥ 2. Then, note that a² + 1/a² = t² - 2. So √(a² + 1/a²) = √(t² - 2). So the left side becomes √(t² - 2) + 2, and the right side is t + √2. Therefore, the inequality becomes:√(t² - 2) + 2 ≥ t + √2.So now, substituting t ≥ 2, need to show this inequality holds. Let me check t = 2. Then left side is √(4 - 2) + 2 = √2 + 2, and right side is 2 + √2. So equality holds here as before. Now, maybe we can consider t ≥ 2 and prove that √(t² - 2) + 2 ≥ t + √2.Let me rearrange this inequality:√(t² - 2) - t ≥ √2 - 2.But the left side is √(t² - 2) - t. Let me compute this. Let's factor √(t² - 2) - t:Multiply numerator and denominator by √(t² - 2) + t:[√(t² - 2) - t][√(t² - 2) + t] / [√(t² - 2) + t] = (t² - 2 - t²) / [√(t² - 2) + t] = (-2)/[√(t² - 2) + t]So √(t² - 2) - t = -2/[√(t² - 2) + t]So the inequality becomes:-2/[√(t² - 2) + t] ≥ √2 - 2Multiply both sides by [√(t² - 2) + t]. But I need to be careful with the direction of the inequality since [√(t² - 2) + t] is positive (since t ≥ 2). So multiplying both sides by a positive number doesn't change the inequality direction. So:-2 ≥ (√2 - 2)[√(t² - 2) + t]Let me compute (√2 - 2)[√(t² - 2) + t]First, note that √2 - 2 is negative since √2 ≈1.414, so √2 - 2 ≈ -0.586. So we have a negative number multiplied by a positive number, which gives a negative number. So the right-hand side is negative, and the left-hand side is -2. So the inequality is:-2 ≥ (√2 - 2)(√(t² - 2) + t)Let me rearrange this. Since both sides are negative, if I take reciprocals, the inequality flips. Alternatively, maybe multiply both sides by -1, which reverses the inequality:2 ≤ (2 - √2)(√(t² - 2) + t)So 2 ≤ (2 - √2)(√(t² - 2) + t)Hmm. Let me compute (2 - √2)(√(t² - 2) + t). Let's see if this is greater than or equal to 2.Alternatively, maybe this approach is getting too convoluted. Let me try another way.Original inequality after substitution: √(t² - 2) + 2 ≥ t + √2.Let me bring all terms to the left side:√(t² - 2) + 2 - t - √2 ≥ 0.So, need to show that √(t² - 2) - t + (2 - √2) ≥ 0.Hmm. Let me consider the function f(t) = √(t² - 2) - t + (2 - √2). I need to show that f(t) ≥ 0 for t ≥ 2.Let me compute f(2): √(4 - 2) - 2 + (2 - √2) = √2 - 2 + 2 - √2 = 0. So equality holds here.Now, let's check the derivative of f(t) to see if the function is increasing or decreasing. If the function is increasing for t ≥ 2, then since f(2) = 0, and t ≥ 2, then f(t) ≥ 0. If the derivative is decreasing, then we need to check the limit as t approaches infinity.Compute f'(t):d/dt [√(t² - 2)] - 1 + 0 = ( (1/(2√(t² - 2)) ) * 2t ) - 1 = t / √(t² - 2) - 1.So f'(t) = t / √(t² - 2) - 1.Let me see if this derivative is positive or negative for t > 2.Compute t / √(t² - 2):Multiply numerator and denominator by 1/t:1 / √(1 - 2/t²)Since t > 2, 2/t² < 0.5, so √(1 - 2/t²) < 1. Therefore, 1 / √(1 - 2/t²) > 1. Hence, t / √(t² - 2) > 1. Therefore, f'(t) = t / √(t² - 2) - 1 > 0.So f(t) is increasing for t > 2. Since f(2) = 0 and f'(t) > 0 for t > 2, then f(t) ≥ 0 for all t ≥ 2. Therefore, the original inequality holds.Wait, that seems to work. Let me verify this with an example. Take a = 2, so t = 2 + 1/2 = 2.5. Then, left side: √(4 + 1/4) + 2 = √(17/4) + 2 = (√17)/2 + 2 ≈ 4.123/2 + 2 ≈ 2.0615 + 2 ≈ 4.0615. Right side: 2.5 + √2 ≈ 2.5 + 1.414 ≈ 3.914. So 4.0615 ≥ 3.914, which holds.Another example: a approaching infinity. Let a = 1000, so t = 1000 + 1/1000 ≈ 1000.001. Then left side: √(1000² + (1/1000)^2) + 2 ≈ √(1000000) + 2 ≈ 1000 + 2 = 1002. Right side: 1000.001 + √2 ≈ 1000.001 + 1.414 ≈ 1001.415. So 1002 ≥ 1001.415, which holds.Another case: a approaching 0. Let a = 0.001, then t = 0.001 + 1000 ≈ 1000.001, same as above. Left side is same as above, right side same. So still holds.Therefore, since after substituting t = a + 1/a, and analyzing the function f(t) = √(t² - 2) + 2 - t - √2, we saw that f(t) is increasing for t ≥ 2 and f(2) = 0, hence f(t) ≥ 0 for all t ≥ 2. Therefore, the inequality holds for all a > 0.Alternatively, another approach could be using the Cauchy-Schwarz inequality or other inequalities, but the substitution seems to have worked. Let me see if another method is possible.Alternatively, consider the left side: √(a² + 1/a²) + 2. Let me denote x = a + 1/a. Then, as before, x ≥ 2, and a² + 1/a² = x² - 2. So the left side becomes √(x² - 2) + 2. The right side is x + √2. So the inequality is √(x² - 2) + 2 ≥ x + √2.Alternatively, rearrange terms:√(x² - 2) - x ≥ √2 - 2.But as before, the left side is negative, but since √2 - 2 is also negative, maybe we can compare them. But I think the earlier approach with the derivative is solid.Alternatively, square both sides. Let me try that.Starting with √(x² - 2) + 2 ≥ x + √2.Subtract 2 from both sides: √(x² - 2) ≥ x + √2 - 2.Now, the left side is √(x² - 2), which is real since x ≥ 2. The right side is x + √2 - 2. Let's see if the right side is positive. For x ≥ 2, x + √2 - 2 ≥ 2 + √2 - 2 = √2 > 0. So both sides are positive, so squaring is okay.Square both sides:x² - 2 ≥ (x + √2 - 2)².Expand the right side:(x + (√2 - 2))² = x² + 2x(√2 - 2) + (√2 - 2)².So:x² - 2 ≥ x² + 2x(√2 - 2) + (√2 - 2)².Subtract x² from both sides:-2 ≥ 2x(√2 - 2) + (√2 - 2)².Let me compute (√2 - 2)² = 2 - 4√2 + 4 = 6 - 4√2.So the inequality becomes:-2 ≥ 2x(√2 - 2) + 6 - 4√2.Bring all terms to the left side:-2 - 6 + 4√2 ≥ 2x(√2 - 2).Simplify left side: -8 + 4√2.Divide both sides by 2 (since 2 is positive, inequality direction remains):(-8 + 4√2)/2 ≥ x(√2 - 2).Which is (-4 + 2√2) ≥ x(√2 - 2).Note that √2 - 2 is negative, so multiply both sides by (√2 - 2), inequality reverses:(-4 + 2√2)/(√2 - 2) ≤ x.Compute (-4 + 2√2)/(√2 - 2). Let's rationalize the denominator:Multiply numerator and denominator by (√2 + 2):Numerator: (-4 + 2√2)(√2 + 2) = -4√2 -8 + 2*2 + 4√2 = (-4√2 + 4√2) + (-8 + 4) = 0 - 4 = -4.Denominator: (√2 - 2)(√2 + 2) = 2 - 4 = -2.So (-4)/(-2) = 2. Therefore:2 ≤ x.But x = a + 1/a ≥ 2 by AM ≥ GM. So 2 ≤ x is always true. Therefore, the inequality reduces to 2 ≤ x, which is true. Hence, the original squared inequality holds, so the original inequality holds.Therefore, squaring both sides and simplifying leads us to an inequality that is always true given x ≥ 2, hence proving the original inequality.Therefore, both methods (using calculus and substitution, and squaring both sides) confirm that the inequality holds for all a > 0.Final AnswerThe inequality is proven to hold for all ( a > 0 ), so the final result is boxed{sqrt{a^2 + frac{1}{a^2}} + 2 geq a + frac{1}{a} + sqrt{2}}.</think>To prove the inequality (sqrt{a^2 + frac{1}{a^2}} + 2 geq a + frac{1}{a} + sqrt{2}) for (a > 0), we can use substitution and algebraic manipulation.1. Substitute (t = a + frac{1}{a}): - By the AM-GM inequality, (t geq 2). - Note that (a^2 + frac{1}{a^2} = t^2 - 2).2. Rewrite the inequality using (t): - The left side becomes (sqrt{t^2 - 2} + 2). - The right side becomes (t + sqrt{2}). - The inequality transforms to (sqrt{t^2 - 2} + 2 geq t + sqrt{2}).3. Rearrange the inequality: - Subtract (t) and (sqrt{2}) from both sides: (sqrt{t^2 - 2} - t + 2 - sqrt{2} geq 0).4. Analyze the function (f(t) = sqrt{t^2 - 2} - t + 2 - sqrt{2}): - Check (f(2)): (sqrt{4 - 2} - 2 + 2 - sqrt{2} = sqrt{2} - 2 + 2 - sqrt{2} = 0). - Compute the derivative (f'(t)): (frac{t}{sqrt{t^2 - 2}} - 1). - Since (t geq 2), (frac{t}{sqrt{t^2 - 2}} > 1), hence (f'(t) > 0), meaning (f(t)) is increasing for (t geq 2).5. Conclusion from the analysis: - Since (f(2) = 0) and (f(t)) is increasing for (t geq 2), (f(t) geq 0) for all (t geq 2).6. Verification by squaring both sides: - Starting with (sqrt{t^2 - 2} + 2 geq t + sqrt{2}), square both sides and simplify. - The resulting inequality reduces to (2 leq t), which is always true since (t geq 2).Thus, the inequality (sqrt{a^2 + frac{1}{a^2}} + 2 geq a + frac{1}{a} + sqrt{2}) holds for all (a > 0).[boxed{sqrt{a^2 + frac{1}{a^2}} + 2 geq a + frac{1}{a} + sqrt{2}}]]

🔑:To prove sqrt{a^2 + frac{1}{a^2}} + 2 geq a + frac{1}{a} + sqrt{2}:Since a > 0, both sides of the inequality are greater than zero. Therefore, we need to show left(sqrt{a^2 + frac{1}{a^2}} + 2right)^2 geq left(a + frac{1}{a} + sqrt{2}right)^2. This is because the square of a positive number retains the inequality sign.Expand both sides of the inequality:left( a^2 + frac{1}{a^2} + 2sqrt{a^2 + frac{1}{a^2}} + 4 right) geq left( a^2 + 2afrac{1}{a} + frac{1}{a^2} + 2asqrt{2} + 2frac{1}{a}sqrt{2} + 2 right).Simplify the left side:a^2 + frac{1}{a^2} + 4 + 2sqrt{a^2 + frac{1}{a^2}} geq a^2 + frac{1}{a^2} + 2 + 2sqrt{2}left(a+frac{1}{a}right).Subtract a^2 + frac{1}{a^2} + 2 from both sides:2 + 2sqrt{a^2 + frac{1}{a^2}} geq 2sqrt{2}left(a+frac{1}{a}right).Divide both sides by 2:1 + sqrt{a^2 + frac{1}{a^2}} geq sqrt{2}left(a+frac{1}{a}right).At this point, we need to prove:sqrt{a^2 + frac{1}{a^2}} geq sqrt{2}left(a + frac{1}{a} right) - 1.This is equivalent to proving:a^2 + frac{1}{a^2} geq 2left(a + frac{1}{a} right)^2 - 2sqrt{2}left(a + frac{1}{a}right) + 1.By AM-GM inequality, we have:a^2 + frac{1}{a^2} geq 2.It follows that:a^2 + frac{1}{a^2} + 4 geq 2left(a + frac{1}{a} right)^2 + 2.Thus, removing the 2 on both sides, we end up with the required inequality:a^2 + frac{1}{a^2} + 2 geq left(a + frac{1}{a} right)^2.This completes the proof. Hence, the original inequality holds true:boxed{sqrt{a^2 + frac{1}{a^2}} + 2 geq a + frac{1}{a} + sqrt{2}}.

🧠:[<think>Okay, let's see. I need to solve this algebraic expression: [(x + 3y)² - (x + y)(x - y)] divided by 2y. Hmm, that looks a bit complicated, but maybe if I break it down step by step, it will make more sense.First, let's tackle the numerator part: (x + 3y)² minus (x + y)(x - y). I remember that expanding squares and products might help simplify things here. Let's start with expanding (x + 3y)². The formula for squaring a binomial is (a + b)² = a² + 2ab + b². Applying that here, where a is x and b is 3y, so:(x + 3y)² = x² + 2*(x)*(3y) + (3y)²= x² + 6xy + 9y²Okay, that part seems done. Now, the next part is subtracting (x + y)(x - y). Wait, that's a difference of squares, right? The formula for (a + b)(a - b) is a² - b². Let me confirm:(x + y)(x - y) = x² - y²Yes, that's correct. So, substituting back into the numerator:Numerator = (x² + 6xy + 9y²) - (x² - y²)Now, let's simplify that. Distribute the negative sign to both terms in the second part:= x² + 6xy + 9y² - x² + y²Ah, the x² and -x² cancel each other out. So:= 6xy + 9y² + y²= 6xy + 10y²So the numerator simplifies to 6xy + 10y². Now, the entire expression is this numerator divided by 2y. So:(6xy + 10y²) ÷ 2yI can factor out a common factor in the numerator. Let's see, both terms have a y, and the coefficients 6 and 10 have a common factor of 2. So factoring out 2y:= 2y(3x + 5y) ÷ 2yThe 2y in the numerator and denominator cancel out:= 3x + 5yWait, that seems too straightforward. Let me check my steps again to make sure I didn't make a mistake.Starting over:1. Expand (x + 3y)²: x² + 6xy + 9y². Correct.2. Expand (x + y)(x - y): x² - y². Correct.3. Subtract the two: (x² + 6xy + 9y²) - (x² - y²) = 6xy + 10y². Yes, x² cancels with -x², and 9y² + y² is 10y².4. Divide by 2y: (6xy + 10y²)/(2y). Factoring out 2y gives 2y(3x + 5y)/2y = 3x + 5y. Yes, that's right.Hmm, maybe another way to check. Let's assign specific values to x and y and see if both the original expression and the simplified expression give the same result.Let's pick x = 1 and y = 1.Original expression:[(1 + 3*1)^2 - (1 + 1)(1 - 1)] / (2*1)= [(4)^2 - (2)(0)] / 2= [16 - 0]/2= 16/2= 8Simplified expression: 3*1 + 5*1 = 3 + 5 = 8. Okay, that matches.Another test: x = 2, y = 3.Original:[(2 + 3*3)^2 - (2 + 3)(2 - 3)] / (2*3)= [(2 + 9)^2 - (5)(-1)] / 6= [11^2 - (-5)] / 6= [121 + 5] /6= 126/6= 21Simplified: 3*2 + 5*3 = 6 + 15 = 21. Correct.One more: x = 0, y = 5.Original:[(0 + 3*5)^2 - (0 +5)(0 -5)] / (2*5)= [(15)^2 - (5)(-5)] /10= [225 +25]/10= 250/10=25Simplified: 3*0 +5*5 =0 +25=25. Correct.So all the test cases work. Therefore, my simplification is correct. The answer is 3x +5y.Final Answerboxed{3x + 5y}</think>To solve the given expression ( left[(x + 3y)^2 - (x + y)(x - y)right] div 2y ), we proceed as follows:1. Expand ((x + 3y)^2): [ (x + 3y)^2 = x^2 + 2(x)(3y) + (3y)^2 = x^2 + 6xy + 9y^2 ]2. Expand ((x + y)(x - y)) using the difference of squares formula: [ (x + y)(x - y) = x^2 - y^2 ]3. Subtract the expanded forms: [ (x^2 + 6xy + 9y^2) - (x^2 - y^2) = x^2 + 6xy + 9y^2 - x^2 + y^2 = 6xy + 10y^2 ]4. Divide the result by (2y): [ frac{6xy + 10y^2}{2y} = frac{2y(3x + 5y)}{2y} = 3x + 5y ]Thus, the final answer is (boxed{3x + 5y}).,<think>Okay, let me try to figure out this algebra problem. The question is: [(x + 3y)² - (x + y)(x - y)] divided by 2y. Hmm, so I need to simplify the expression inside the brackets first, right? Then divide the result by 2y. Let me break it down step by step.First, let's expand (x + 3y)². I remember that squaring a binomial follows the formula (a + b)² = a² + 2ab + b². Applying that here, a would be x and b is 3y. So, expanding that:(x + 3y)² = x² + 2*x*(3y) + (3y)²= x² + 6xy + 9y²Okay, got that part. Now, the next term is (x + y)(x - y). That's a product of a binomial and its conjugate, which is a difference of squares. The formula is (a + b)(a - b) = a² - b². Here, a is x and b is y. So:(x + y)(x - y) = x² - y²Alright, so now the original expression inside the brackets is:(x + 3y)² - (x + y)(x - y) = [x² + 6xy + 9y²] - [x² - y²]Let me subtract the second expression from the first. Distribute the negative sign to both terms in the second bracket:= x² + 6xy + 9y² - x² + y²Now, let's combine like terms. The x² terms: x² - x² cancels out. Then we have 6xy. For the y² terms: 9y² + y² = 10y². So, combining them:= 6xy + 10y²So the numerator simplifies to 6xy + 10y². Now, the entire expression is this divided by 2y. So:(6xy + 10y²) ÷ 2yI can factor out a common factor from the numerator. Let's see, both terms have a factor of 2y. Let me check:6xy = 2y * 3x10y² = 2y * 5ySo factoring 2y out:= 2y(3x + 5y) ÷ 2yNow, the 2y in the numerator and denominator cancels out, assuming y ≠ 0 (since division by zero is undefined). So we're left with:3x + 5yTherefore, the simplified expression is 3x + 5y.Wait, let me double-check each step to make sure I didn't make any mistakes. First, expanding (x + 3y)²: x², 2*x*3y is 6xy, and (3y)² is 9y². That seems correct. Then (x + y)(x - y) is x² - y². Subtracting that from the first expansion: so x² + 6xy + 9y² - x² + y². The x² cancels, 6xy remains, and 9y² + y² is 10y². Then factoring 2y from 6xy + 10y², which gives 2y(3x + 5y). Divided by 2y cancels to 3x + 5y. Looks good.Alternatively, another way to approach the division is to split the terms:(6xy + 10y²) / 2y = (6xy)/2y + (10y²)/2ySimplify each term:6xy ÷ 2y = (6/2)(x)(y/y) = 3x*1 = 3x10y² ÷ 2y = (10/2)(y²/y) = 5ySo adding them together gives 3x + 5y. Yep, same result. That confirms my answer.I think that's correct. Let me just verify once more. If I substitute actual numbers for x and y, compute the original expression and check if it equals 3x + 5y.Let's choose x = 1, y = 1.Original expression:[(1 + 3*1)² - (1 + 1)(1 - 1)] ÷ (2*1)First, compute inside the brackets:(1 + 3)² - (2)(0) = (4)² - 0 = 16 - 0 = 16Divide by 2*1 = 2: 16 ÷ 2 = 8Now, compute 3x + 5y with x=1, y=1: 3*1 + 5*1 = 8. Same result. Good.Another test with x=2, y=3.Original expression:[(2 + 3*3)² - (2 + 3)(2 - 3)] ÷ (2*3)Compute inside brackets:(2 + 9)² - (5)(-1) = (11)² - (-5) = 121 + 5 = 126Divide by 6: 126 ÷ 6 = 213x + 5y: 3*2 + 5*3 = 6 + 15 = 21. Correct again.Looks like the answer is correct. So I'm confident that the simplified expression is 3x + 5y.Final AnswerThe simplified expression is boxed{3x + 5y}.</think>First, expand ((x + 3y)^2) using the binomial expansion formula:[(x + 3y)^2 = x^2 + 2 cdot x cdot 3y + (3y)^2 = x^2 + 6xy + 9y^2]Next, expand ((x + y)(x - y)) using the difference of squares formula:[(x + y)(x - y) = x^2 - y^2]Subtract the second expansion from the first:[(x^2 + 6xy + 9y^2) - (x^2 - y^2) = x^2 + 6xy + 9y^2 - x^2 + y^2 = 6xy + 10y^2]Divide the result by (2y):[frac{6xy + 10y^2}{2y} = frac{2y(3x + 5y)}{2y} = 3x + 5y]Thus, the simplified expression is (boxed{3x + 5y}).]

🔑:To solve the given problem, we start with the original expression and simplify it step by step:[begin{align*}left[left(x+3yright)^{2}-left(x+yright)left(x-yright)right]div 2y &= left[(x^{2}+6xy+9y^{2})-(x^{2}-y^{2})right]div 2y &= left[x^{2}+6xy+9y^{2}-x^{2}+y^{2}right]div 2y &= left[6xy+10y^{2}right]div 2y &= 3x+5y.end{align*}]Therefore, the simplified expression is boxed{3x+5y}.