🧠:[<think>Okay, so I need to solve this problem where a line passes through the point (10, 0) and intersects the parabola y = x² at two points with abscissas x₁ and x₂. Then I have to find 1/x₁ + 1/x₂. Hmm, abscissas mean the x-coordinates, right? So x₁ and x₂ are the x-values where the line intersects the parabola. Got it.First, let me visualize this. The parabola y = x² is a standard upward-opening parabola. The line passes through (10, 0), which is a point on the x-axis, 10 units to the right of the origin. The line intersects the parabola at two points, so it's a secant line to the parabola. The x-coordinates of these intersection points are x₁ and x₂, and I need to find the sum of their reciprocals.To approach this, I remember that if a line intersects a parabola, the points of intersection can be found by solving the system of equations consisting of the line and the parabola. So first, I need to find the equation of the line that passes through (10, 0) and intersects the parabola at x₁ and x₂. But wait, the problem doesn't specify which line—it could be any line passing through (10, 0). However, the problem states that the line intersects the parabola at two points, so the line is not vertical (since vertical lines would have an undefined slope and only intersect the parabola once if at all). So the line must have some slope m, and pass through (10, 0).Let me denote the equation of the line in point-slope form. Since it passes through (10, 0), the equation would be:y - 0 = m(x - 10)=> y = m(x - 10)So the equation of the line is y = m(x - 10). Now, to find the points where this line intersects the parabola y = x², I need to set them equal:x² = m(x - 10)That gives the quadratic equation:x² - m x + 10 m = 0Wait, let's check that. If I move everything to one side:x² - m x + 10 m = 0Yes. So this quadratic equation has solutions x₁ and x₂, which are the abscissas (x-coordinates) of the intersection points.Now, since we're dealing with a quadratic equation, I can use Vieta's formulas to find the sum and product of the roots. Vieta's says that for a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a and the product is c/a.In this case, the quadratic equation is x² - m x + 10 m = 0. So a = 1, b = -m, c = 10m.Therefore:x₁ + x₂ = -b/a = -(-m)/1 = mx₁ x₂ = c/a = 10m / 1 = 10mOkay, so the sum of the roots x₁ + x₂ is m, and the product x₁ x₂ is 10m.But the problem is asking for 1/x₁ + 1/x₂. Let me express that in terms of the sum and product of the roots. 1/x₁ + 1/x₂ = (x₂ + x₁)/(x₁ x₂) = (x₁ + x₂)/(x₁ x₂)From Vieta's formulas, we have x₁ + x₂ = m and x₁ x₂ = 10m. Therefore:1/x₁ + 1/x₂ = m / (10m) = 1/10Wait, that simplifies to 1/10? That's interesting. So regardless of the slope m, the value of 1/x₁ + 1/x₂ is always 1/10? That seems counterintuitive. Let me check my steps again.First, equation of the line: y = m(x - 10). Correct.Set equal to parabola: x² = m(x - 10) => x² - m x + 10m = 0. Correct.Coefficients: a = 1, b = -m, c = 10m. So sum of roots is m, product is 10m. Correct.Then 1/x₁ + 1/x₂ = (x₁ + x₂)/(x₁ x₂) = m / (10m) = 1/10. Yes, m cancels out. So the answer is 1/10.But wait, does this hold for all lines passing through (10, 0) except the vertical line? Because vertical line would be x = 10, which intersects the parabola at x = 10, y = 100. But since vertical lines aren't considered here (they have undefined slope and only one intersection point if at all), so all non-vertical lines through (10,0) will result in 1/x₁ + 1/x₂ = 1/10. So the answer is 1/10.Alternatively, maybe I can test this with a specific example. Let's take a specific line through (10, 0) and compute 1/x₁ + 1/x₂.For example, suppose the line is horizontal. But wait, a horizontal line through (10, 0) would be y = 0. Then intersection with parabola y = x² is at x = 0. But that's only one point. But the problem states that the line intersects the parabola at two points, so horizontal line y = 0 is tangent to the parabola at the origin. So maybe in that case, the line is tangent, so x₁ = x₂ = 0. Then 1/x₁ + 1/x₂ would be undefined. But in the problem, they mention points with abscissas x₁ and x₂, implying two distinct points. So maybe the line is not tangent, so we can assume m ≠ 0? Wait, but even if the line is horizontal, m = 0, but that line only intersects at x = 0, which is a double root. So perhaps the problem is considering lines that intersect at two distinct points, so m is such that the quadratic has two distinct roots, i.e., discriminant is positive.But even so, in my previous reasoning, the answer 1/10 does not depend on m. So even if m approaches zero, making the line almost horizontal, the sum 1/x₁ + 1/x₂ would approach 1/10. But if m = 0, then the line is horizontal, intersects at x = 0 (double root), but in that case, 1/x₁ + 1/x₂ would be undefined (since both x₁ and x₂ are zero). So perhaps the problem assumes that the line is not tangent, so the quadratic equation has two distinct roots, so discriminant must be positive.But in our case, discriminant D = b² - 4ac = (-m)² - 4*1*10m = m² - 40m. For D > 0, m² - 40m > 0 => m(m - 40) > 0. So either m < 0 or m > 40. So for lines with slope m < 0 or m > 40, the line intersects the parabola at two distinct points.But regardless, even with that, the expression 1/x₁ + 1/x₂ = 1/10 holds for any such m. Let's pick m = 50 (which is greater than 40, so discriminant D = 2500 - 2000 = 500 > 0). Then the quadratic equation is x² -50x + 500 = 0. Solving this, x = [50 ± sqrt(2500 - 2000)] / 2 = [50 ± sqrt(500)] / 2 = [50 ± 10*sqrt(5)] / 2 = 25 ± 5*sqrt(5). Then x₁ = 25 + 5√5, x₂ = 25 - 5√5. Then 1/x₁ + 1/x₂ = [ (25 - 5√5) + (25 + 5√5) ] / [ (25 + 5√5)(25 - 5√5) ] = (50) / (625 - 125) = 50 / 500 = 1/10. So that works.Another example: take m = -10 (which is less than 0). Then the quadratic equation is x² - (-10)x + 10*(-10) = x² +10x -100 = 0. Solutions: x = [-10 ± sqrt(100 + 400)] / 2 = [-10 ± sqrt(500)] / 2 = [-10 ± 10√5]/2 = -5 ± 5√5. Then x₁ = -5 +5√5, x₂ = -5 -5√5. Then 1/x₁ +1/x₂ = [ (-5 -5√5) + (-5 +5√5) ] / [ (-5 +5√5)(-5 -5√5) ] = (-10)/[25 - (5√5)^2] = (-10)/[25 - 125] = (-10)/(-100) = 1/10. So that also works.Even if the line is vertical, but vertical line x=10 intersects the parabola at x=10, but that's only one point. So vertical line is excluded.Therefore, regardless of the slope (as long as the line intersects the parabola at two distinct points), the value of 1/x₁ +1/x₂ is always 1/10.Hence, the answer is 1/10, so boxed answer is boxed{dfrac{1}{10}}.Final Answerboxed{dfrac{1}{10}}</think>A line passes through the point ((10, 0)) and intersects the parabola (y = x^2) at points with abscissas (x_1) and (x_2). We need to find (frac{1}{x_1} + frac{1}{x_2}).1. The equation of the line passing through ((10, 0)) can be written in point-slope form as (y = m(x - 10)), where (m) is the slope.2. Setting this equal to the parabola's equation (y = x^2), we get: [ x^2 = m(x - 10) ]3. Rearranging into a standard quadratic equation form: [ x^2 - mx + 10m = 0 ]4. Using Vieta's formulas for the quadratic equation (x^2 - mx + 10m = 0), the sum and product of the roots (x_1) and (x_2) are: [ x_1 + x_2 = m quad text{and} quad x_1 x_2 = 10m ]5. The expression (frac{1}{x_1} + frac{1}{x_2}) can be rewritten using the sum and product of the roots: [ frac{1}{x_1} + frac{1}{x_2} = frac{x_1 + x_2}{x_1 x_2} = frac{m}{10m} = frac{1}{10} ]Thus, the final answer is (boxed{dfrac{1}{10}}).,<think>Okay, let me see. So the problem is about a line passing through the point (10, 0) and intersecting the parabola y = x² at two points with abscissas x₁ and x₂. We need to find 1/x₁ + 1/x₂. Hmm, abscissas are the x-coordinates, right? So the line intersects the parabola at (x₁, x₁²) and (x₂, x₂²). The question is asking for the sum of the reciprocals of these x-coordinates.First, let me recall that when a line intersects a parabola, the points of intersection can be found by solving the equations simultaneously. The equation of the line isn't given directly, but we know it passes through (10, 0). So maybe I can write the equation of the line in a general form and then use the fact that it passes through (10, 0) to find a relationship between the coefficients.Let me start by assuming the equation of the line is y = mx + c. Since it passes through (10, 0), substituting x = 10 and y = 0 into the equation gives 0 = 10m + c, so c = -10m. Therefore, the equation of the line can be written as y = mx - 10m.Now, to find the points of intersection between this line and the parabola y = x², I can set them equal:x² = mx - 10m.Let me rearrange this equation to standard quadratic form:x² - mx + 10m = 0.So, the quadratic equation is x² - mx + 10m = 0. The solutions to this equation are the x-coordinates x₁ and x₂ where the line intersects the parabola. Since we need 1/x₁ + 1/x₂, maybe I can use the properties of quadratic equations, specifically Vieta's formulas, which relate the sum and product of roots to the coefficients.Vieta's formulas state that for a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a and the product is c/a. In this case, the equation is x² - mx + 10m = 0, so a = 1, b = -m, and c = 10m. Therefore, the sum of the roots x₁ + x₂ = -b/a = m/1 = m, and the product x₁x₂ = c/a = 10m/1 = 10m.Now, the expression we need is 1/x₁ + 1/x₂. Let's compute that:1/x₁ + 1/x₂ = (x₁ + x₂)/(x₁x₂).Substituting the values from Vieta's formulas, this becomes m/(10m) = 1/10. Wait, that simplifies to 1/10? So regardless of the value of m, as long as the line passes through (10, 0) and intersects the parabola, the sum 1/x₁ + 1/x₂ is always 1/10?But let me check if this is correct. Maybe I made a mistake in assuming the equation of the line. The line passes through (10, 0), so I wrote it as y = m(x - 10). Then, intersecting with y = x² gives x² = m(x - 10). Then moving everything to one side, x² - mx + 10m = 0. Vieta's formulas: sum is m, product is 10m, so indeed 1/x₁ + 1/x₂ = (x₁ + x₂)/(x₁x₂) = m/(10m) = 1/10. So yes, the m cancels out. Therefore, the answer is 1/10.But let me think again. What if the line is vertical? A vertical line passing through (10, 0) would have the equation x = 10. But intersecting with the parabola y = x² would give the point (10, 100). But that's only one point, so there's only one intersection. However, the problem states that the line intersects the parabola at two points, so the vertical line isn't considered here because it's tangent or only intersects once? Wait, actually, a vertical line x = 10 intersects the parabola y = x² at (10, 100). So that's just one point. So the problem says the line intersects the parabola at two points, so vertical lines are excluded. Therefore, the line must have a slope, so m is defined, and we can proceed with the previous reasoning.Alternatively, maybe using parametric equations or another approach, but this seems straightforward.Wait, but another way to check is to take a specific line passing through (10, 0) and compute 1/x₁ + 1/x₂. Let's pick a specific line. For example, take m = 1. Then the line is y = x - 10. Setting this equal to y = x² gives x² = x - 10 => x² - x + 10 = 0. The discriminant is 1 - 40 = -39, which is negative. So no real solutions. Wait, that's a problem. So if m = 1, there are no real intersections. So in that case, the line does not intersect the parabola. But the problem states that the line intersects the parabola at two points. So m cannot be arbitrary. So maybe m has to be such that the discriminant is positive.So discriminant D = (-m)^2 - 4*1*10m = m² - 40m. For real roots, D ≥ 0, so m² - 40m ≥ 0. Therefore, m(m - 40) ≥ 0. So m ≤ 0 or m ≥ 40. So lines with slope m ≤ 0 (i.e., horizontal or decreasing) or m ≥ 40 (steeply increasing) will intersect the parabola at two points.But in our previous calculation, we found that 1/x₁ + 1/x₂ = 1/10, regardless of m. Let's test with a valid m. Let's take m = 40. Then the line is y = 40x - 400. Setting equal to y = x² gives x² - 40x + 400 = 0. The discriminant is 1600 - 1600 = 0. So one real root, but the problem states two points. Hmm, so m = 40 gives a tangent line. So maybe m has to be greater than 40 or less than 0. Let me take m = -1. Then the line is y = -x + 10. Intersection with parabola: x² = -x + 10 => x² + x -10 = 0. Discriminant is 1 + 40 = 41, which is positive. So two real roots. Let's compute them. x = [-1 ± sqrt(41)]/2. Then 1/x₁ + 1/x₂ = (x₁ + x₂)/(x₁x₂). From equation x² + x -10 =0, sum of roots is -1/1 = -1, product is -10/1 = -10. Therefore, 1/x₁ + 1/x₂ = (-1)/(-10) = 1/10. So indeed, it works.Another example: take m = 50. Line is y = 50x - 500. Intersection with parabola: x² = 50x - 500 => x² -50x + 500 = 0. Discriminant: 2500 - 2000 = 500 > 0. So two real roots. Sum of roots x₁ + x₂ = 50, product x₁x₂ = 500. Then 1/x₁ + 1/x₂ = 50/500 = 1/10. Correct again.Even when m is negative, as in the previous case, we got 1/10. So the answer seems to hold. Therefore, regardless of the slope (as long as the line intersects the parabola at two distinct points), 1/x₁ + 1/x₂ is always 1/10. Therefore, the answer is 1/10.Final AnswerThe value of frac{1}{x_{1}}+frac{1}{x_{2}} is boxed{dfrac{1}{10}}.</think>Given a line passing through the point ((10, 0)) and intersecting the parabola (y = x^2) at points with abscissas (x_1) and (x_2), we need to find (frac{1}{x_1} + frac{1}{x_2}).1. Equation of the Line: The line passes through ((10, 0)), so its equation can be written as (y = m(x - 10)), where (m) is the slope.2. Intersection with Parabola: Setting the equation of the line equal to the parabola (y = x^2), we get: [ x^2 = m(x - 10) ] Rearranging this into standard quadratic form: [ x^2 - mx + 10m = 0 ]3. Vieta's Formulas: For the quadratic equation (x^2 - mx + 10m = 0), the sum of the roots (x_1 + x_2 = m) and the product of the roots (x_1x_2 = 10m).4. Expression Calculation: Using Vieta's formulas, we compute (frac{1}{x_1} + frac{1}{x_2}): [ frac{1}{x_1} + frac{1}{x_2} = frac{x_1 + x_2}{x_1x_2} = frac{m}{10m} = frac{1}{10} ]Thus, the value of (frac{1}{x_1} + frac{1}{x_2}) is (boxed{dfrac{1}{10}}).]

🔑:Answer: 0.1.Solution. The points mathrm{x}_{1}, mathrm{x}_{2}, are solutions to the equation x^{2}=k(x-10), where mathrm{k} is the slope coefficient of the line. Then, by Vieta's theorem, x_{1}+x_{2}=k, x_{1} x_{2}=10 k. Therefore, frac{1}{x_{1}}+frac{1}{x_{2}}=frac{x_{1}+x_{2}}{x_{1} x_{2}}=frac{k}{10 k}=frac{1}{10}.

🧠:[<think>Okay, let's see. The problem is to represent 2/7 as the sum of several different common fractions where each numerator is one. So, these are unit fractions, right? Like 1/2, 1/3, 1/4, etc. And they all have to be different. Hmm, interesting. I remember that Egyptians used to do this kind of thing with their fractions. Maybe there's a method to break down fractions like this.First, I should recall if there's a standard algorithm for expressing fractions as sums of distinct unit fractions. I think one method is the greedy algorithm. Let me try that. The idea is to subtract the largest possible unit fraction from the original fraction and then repeat the process with the remainder.So starting with 2/7. What's the largest unit fraction less than or equal to 2/7? Let's see. The denominators would be integers greater than or equal to (7/2) which is 3.5. So the next integer is 4. So 1/4 is 0.25, and 2/7 is approximately 0.2857. So 1/4 is less than 2/7. Let's subtract 1/4 from 2/7.2/7 - 1/4. To subtract these, find a common denominator, which is 28. 2/7 is 8/28 and 1/4 is 7/28. So 8/28 - 7/28 = 1/28. So now we have 2/7 = 1/4 + 1/28. But is that correct? Let me check. 1/4 is 7/28, and 7/28 + 1/28 is 8/28, which is 2/7. Yes. But the problem says "several different" fractions, so we need more than one. Wait, but 1/4 and 1/28 are two different unit fractions, so maybe that's the answer? But the problem says "several", which could mean two or more. Let me check the problem statement again: "the sum of several different common fractions, the numerators of which are equal to one." So if two are allowed, then this works. But maybe the problem expects more than two? Let me verify.Wait, in Egyptian fractions, they often express fractions as sums of distinct unit fractions. For example, 2/3 would be 1/2 + 1/6. So in this case, maybe 2/7 can be done with two terms. However, when I subtract 1/4 from 2/7, the remainder is 1/28, which is a unit fraction, so that's a valid decomposition. So 2/7 = 1/4 + 1/28. But let me check if there's another way with more terms. Maybe the problem expects more terms? Hmm.Alternatively, maybe the greedy algorithm isn't the only way. Let me see. Let's try another approach. Suppose we try to express 2/7 as the sum of three unit fractions. Let's denote them as 1/a + 1/b + 1/c where a < b < c. Then we need 1/a + 1/b + 1/c = 2/7. Let's try to find such a, b, c.Since all denominators are distinct and greater than 1. Let's try starting with the smallest possible denominators. Let's start with 1/3. Then 2/7 - 1/3 is equal to (6/21 - 7/21) = -1/21. That's negative, which is no good. So 1/3 is too big. Next, try 1/4. Then 2/7 -1/4 = 1/28 as before. So if we take 1/4 +1/28, but that's two fractions. If we need three fractions, perhaps we can split 1/28 into two unit fractions. But how?Wait, splitting 1/28 into two unit fractions. Let's see. 1/28 = 1/x +1/y. Then, we need 1/x +1/y =1/28. Let's assume x < y. Then x must be greater than 28. Let's try x=29. Then 1/y =1/28 -1/29 = (29 -28)/(28*29)=1/812. So 1/28=1/29 +1/812. So then 2/7=1/4 +1/29 +1/812. That's three different unit fractions. But the problem didn't specify the minimum number of fractions, just "several". So maybe two is acceptable. But I wonder if two is considered enough here. Let me check again.The problem says "the sum of several different common fractions, the numerators of which are equal to one." Several just means more than one, so two is acceptable. So 1/4 +1/28 would be a valid answer. But perhaps the problem is expecting a more traditional decomposition. Let me check with another approach.Alternatively, maybe using 1/5 as a starting point. 2/7 -1/5 is (10/35 -7/35)=3/35. Then we need to express 3/35 as a sum of unit fractions. Let's try the greedy algorithm again. The largest unit fraction less than 3/35 is 1/12 (since 35/3 ≈11.666, so next integer is 12). 3/35 -1/12 = (36/420 -35/420)=1/420. So 3/35=1/12 +1/420. Therefore, 2/7=1/5 +1/12 +1/420. That's three terms. Hmm, but is 1/420 really necessary? Maybe there's a better decomposition.Alternatively, let's see if 3/35 can be written as 1/12 +1/420. Or maybe 1/14 +1/35? Let's check 1/14 +1/35. Let's compute: 1/14 is 35/490, 1/35 is 14/490. Together, 49/490 =1/10. Not 3/35. Wait, 3/35 is 42/490. So 1/14 +1/35 = 35/490 +14/490=49/490=1/10. Not helpful. Maybe 1/13 + something? Let's try 3/35 -1/13. 3/35 ≈0.0857, 1/13≈0.0769. The difference is approximately 0.0088, which is about 1/113. So 3/35=1/13 +1/113. But 1/113 is not a standard fraction. Maybe not helpful.Alternatively, maybe using 1/6. Let's try 2/7 -1/6. 2/7 is approximately 0.2857, 1/6 is 0.1667. 0.2857 -0.1667≈0.1190, which is approximately 1/8.4. So 1/8 is 0.125. Let's check exactly: 2/7 -1/6 = (12/42 -7/42)=5/42. 5/42. So 5/42 needs to be represented as a sum of unit fractions. Let's apply the greedy algorithm again. The largest unit fraction less than 5/42 is 1/9 (since 42/5=8.4, so next integer is 9). 5/42 -1/9 = (15/126 -14/126)=1/126. Therefore, 5/42=1/9 +1/126. Therefore, 2/7=1/6 +1/9 +1/126. That's three terms.But again, maybe there's a simpler decomposition. Let's see. Wait, 5/42. Is there a way to split this into two unit fractions? Let's try. Suppose 5/42=1/a +1/b. Then 1/a +1/b=5/42. Let's assume a < b. Then 1/a >5/(42*2)=5/84≈0.0595. So a must be less than 84/5≈16.8. Let's check possible a from 9 upwards (since 1/9 is the previous try). Let's see:If a=9: 1/9 +1/b=5/42 =>1/b=5/42 -1/9=15/126 -14/126=1/126 => b=126. So that's the same as before.If a=10: 1/10 +1/b=5/42 =>1/b=5/42 -1/10=25/210 -21/210=4/210=2/105 =>1/b=2/105 => b=52.5, which is not an integer. So discard.a=11:1/11 +1/b=5/42 =>1/b=5/42 -1/11≈0.1190 -0.0909≈0.0281, which is about 1/35.6. Let's compute exactly: 5/42 -1/11= (55 -42)/462=13/462. So 1/b=13/462 =>b=462/13≈35.54. Not integer.a=12:1/12 +1/b=5/42 =>1/b=5/42 -1/12=10/84 -7/84=3/84=1/28. So b=28. So 5/42=1/12 +1/28. Therefore, 2/7=1/6 +1/12 +1/28. That's three terms as well. Wait, but 1/6, 1/12, and 1/28. All different. So this is another decomposition. Let me verify:1/6 is 14/84, 1/12 is 7/84, 1/28 is 3/84. Total is 14+7+3=24/84=2/7. Correct. So this works. So 2/7=1/6 +1/12 +1/28.Alternatively, let's check if there's a way with two terms. So 2/7=1/a +1/b. Let's solve for integers a and b.So 1/a +1/b=2/7. Let's multiply both sides by 7ab: 7b +7a=2ab.Rearranged: 2ab -7a -7b=0. Let's add 49 to both sides: 2ab -7a -7b +49=49. Factor the left side: (2a -7)(2b -7)=49.Because 2ab -7a -7b +49 = (2a -7)(2b -7). Let me verify:(2a -7)(2b -7) =4ab -14a -14b +49. But we have 2ab -7a -7b +49. So that's not equal. Hmm, maybe another approach.Alternatively, let's use Simon's Favorite Factoring Trick. Start from 2ab -7a -7b =0. Let's add 49/2 to both sides. But maybe this is messy. Alternatively, rearrange:2ab -7a -7b=0Let me write this as:2ab =7a +7bDivide both sides by ab:2 =7/b +7/aLet’s let x=1/a and y=1/b, then:x + y =2/7and 7x +7y=2.But maybe not helpful. Alternatively, let's fix a and solve for b.From 1/a +1/b=2/7.Let's solve for b:1/b=2/7 -1/a = (2a -7)/7aThus,b=7a/(2a -7)Since b must be a positive integer, 2a -7 must divide 7a, and 2a -7>0 => a>3.5. So a≥4.Let’s check possible integer values of a starting from 4:a=4:b=7*4/(8 -7)=28/1=28. So b=28. So 1/4 +1/28=2/7. Which is the first decomposition I found. So yes, a=4, b=28.a=5:b=7*5/(10 -7)=35/3≈11.666. Not integer.a=6:b=42/(12 -7)=42/5=8.4. Not integer.a=7:b=49/(14 -7)=49/7=7. So b=7. But a and b must be different? The problem says "several different common fractions". So if a=7 and b=7, that's 1/7 +1/7=2/7. But they are not different. So invalid.a=8:b=56/(16 -7)=56/9≈6.222. Not integer.a=9:b=63/(18 -7)=63/11≈5.727. Not integer.a=10:b=70/(20 -7)=70/13≈5.384. Not integer.a=11:b=77/(22 -7)=77/15≈5.133. Not integer.a=12:b=84/(24 -7)=84/17≈4.941. Not integer.a=13:b=91/(26 -7)=91/19≈4.789. Not integer.a=14:b=98/(28 -7)=98/21≈4.666. Not integer.a=15:b=105/(30 -7)=105/23≈4.565. Not integer.a=16:b=112/(32 -7)=112/25=4.48. Not integer.a=17:b=119/(34 -7)=119/27≈4.407. Not integer.a=18:b=126/(36 -7)=126/29≈4.344. Not integer.a=19:b=133/(38 -7)=133/31≈4.290. Not integer.a=20:b=140/(40 -7)=140/33≈4.242. Not integer.a=21:b=147/(42 -7)=147/35=4.2. Not integer.a=22:b=154/(44 -7)=154/37≈4.162. Not integer.a=23:b=161/(46 -7)=161/39≈4.128. Not integer.a=24:b=168/(48 -7)=168/41≈4.097. Not integer.a=25:b=175/(50 -7)=175/43≈4.069. Not integer.a=26:b=182/(52 -7)=182/45≈4.044. Not integer.a=27:b=189/(54 -7)=189/47≈4.021. Not integer.a=28:b=196/(56 -7)=196/49=4. So b=4. But a=28, b=4. Since a must be less than b for uniqueness? Wait, we already considered a=4, b=28. So swapping a and b gives the same pair. So the only integer solution is a=4 and b=28. Therefore, the only decomposition into two unit fractions is 1/4 +1/28.So if the problem allows two fractions, then that's the answer. If more are required, then the three-term decompositions like 1/6 +1/12 +1/28 or 1/5 +1/12 +1/420 or 1/4 +1/29 +1/812. But likely, the simplest answer is the two-term decomposition. However, let me check if that's considered valid.But the problem says "several different common fractions", and several usually means more than two. Wait, actually, no. "Several" can mean two or more. So in some contexts, several might mean three or more, but I think in mathematics problems, if they specify "several", two is acceptable. But maybe the problem expects a decomposition with more than two terms. Let me check the problem statement again: "the sum of several different common fractions, the numerators of which are equal to one." There's no specification on the number, just "several", which is vague. But in the example of Egyptian fractions, 2/3 is expressed as 1/2 +1/6, which is two terms. So perhaps two is acceptable here. However, let me check if 1/4 +1/28 is the minimal answer. Alternatively, maybe the problem requires more than two, so let's think of three-term decompositions.Earlier, I found 2/7=1/6 +1/12 +1/28. Let me verify that:1/6 is approximately 0.1667, 1/12≈0.0833, 1/28≈0.0357. Sum≈0.2857, which is 2/7. Correct.Alternatively, 1/4 +1/28 is 0.25 +0.0357≈0.2857. Correct. So both are valid.But perhaps the problem expects the answer with the fewest terms possible. Since two is fewer than three, maybe 1/4 +1/28 is the answer. But perhaps the problem wants all possible decompositions? But the problem says "represent the number 2/7 as the sum...", so maybe any correct decomposition is acceptable. But the problem doesn't specify if multiple answers are possible, so perhaps the simplest one.But to be safe, maybe check if there is another decomposition with two terms. As we saw, the equation 1/a +1/b=2/7 has only one solution with distinct denominators: a=4, b=28. So that's the only two-term decomposition. Therefore, if two is allowed, that's the answer. Otherwise, three terms. But since "several" can mean two, and given that Egyptian fractions often use the minimal number of terms, which would be two here, I think the answer is 1/4 +1/28.But just to be thorough, let me check another three-term decomposition. For example, starting with 1/5:2/7 -1/5 = (10/35 -7/35)=3/35. Now, 3/35. Let's try to split this. Using greedy: the largest unit fraction less than 3/35 is 1/12 (since 35/3≈11.666). So 3/35 -1/12= (36/420 -35/420)=1/420. So 3/35=1/12 +1/420. Therefore, 2/7=1/5 +1/12 +1/420.Alternatively, could we split 3/35 into 1/14 +1/70? Let's check: 1/14 +1/70=5/70 +1/70=6/70=3/35. Yes! So that works. Therefore, 3/35=1/14 +1/70. Therefore, 2/7=1/5 +1/14 +1/70. Let's verify:1/5=14/70, 1/14=5/70, 1/70=1/70. Total:14+5+1=20/70=2/7. Correct. So another three-term decomposition: 1/5 +1/14 +1/70.So there are multiple ways to decompose 2/7 into unit fractions. The problem says "the sum of several different common fractions", so as long as they are different and each has numerator 1, any such decomposition is acceptable. But likely, the answer expected is the one with the fewest terms, which is two: 1/4 +1/28. However, the problem might expect a more "elementary" decomposition, or maybe the one with smaller denominators. Let me see.Comparing denominators in the two-term decomposition: 4 and 28. In the three-term decomposition starting with 1/5: 5,14,70. Alternatively, the three-term decomposition starting with 1/6: 6,12,28. Those denominators are smaller. So perhaps 1/6 +1/12 +1/28 is preferable.Alternatively, let's see which decomposition is more standard. Checking online, but since I can't actually browse, I have to rely on my knowledge. From what I recall, 2/7 in Egyptian fractions is typically represented as 1/4 +1/28. For example, the Rhind Mathematical Papyrus, which is a famous source of Egyptian fractions, represents 2/7 as 1/4 +1/28. So that might be the standard decomposition.Therefore, the answer is likely 1/4 +1/28. However, since the problem didn't specify the number of terms, but says "several", which could be two or more, but given the standard Egyptian fraction decomposition, the two-term version is acceptable. Therefore, the final answer is 1/4 +1/28.But wait, let me check once more. If I follow the greedy algorithm, which takes the largest possible unit fraction each time, then the decomposition is 1/4 +1/28. That's correct. And that is the standard method. So probably that's the answer expected here.</think>To represent (frac{2}{7}) as the sum of several different common fractions with numerator one (unit fractions), we can use the greedy algorithm. This method involves repeatedly subtracting the largest possible unit fraction from the remaining value.1. Find the largest unit fraction less than (frac{2}{7}): - (frac{2}{7} approx 0.2857). The largest unit fraction less than this is (frac{1}{4} = 0.25). 2. Subtract (frac{1}{4}) from (frac{2}{7}): [ frac{2}{7} - frac{1}{4} = frac{8}{28} - frac{7}{28} = frac{1}{28} ] 3. Express the remainder as a unit fraction: - The remainder (frac{1}{28}) is already a unit fraction.Thus, (frac{2}{7}) can be expressed as the sum of two different unit fractions:[frac{2}{7} = frac{1}{4} + frac{1}{28}]Verification:[frac{1}{4} + frac{1}{28} = frac{7}{28} + frac{1}{28} = frac{8}{28} = frac{2}{7}]Therefore, the answer is:[boxed{frac{1}{4} + frac{1}{28}}],<think>Okay, so I need to represent 2/7 as the sum of several different common fractions where each numerator is 1. These kinds of fractions are called unit fractions, right? Like Egyptian fractions. The key points here are that each fraction has to be different, so I can't repeat any unit fractions, and all the numerators are 1.First, I remember that Egyptian fraction expansions typically involve breaking down a fraction into a sum of distinct unit fractions. So maybe there's a method to do that. Let me think. The usual approach is to use the greedy algorithm, where you take the largest possible unit fraction less than the remaining amount and subtract it, then repeat the process. Let me try that here.Starting with 2/7. The largest unit fraction less than 2/7 is 1/4, since 1/4 is 0.25 and 2/7 is approximately 0.2857. So subtract 1/4 from 2/7:2/7 - 1/4 = (8/28 - 7/28) = 1/28.So now, 2/7 = 1/4 + 1/28. Wait, but that's only two fractions. Are they different? Yes, 1/4 and 1/28. But the problem says "several different," and "several" usually means more than two. Hmm. Let me check if the question requires more than two. It says "several," but I'm not sure the exact number. However, maybe 1/4 + 1/28 is sufficient. Let me check if that's correct.Yes, 1/4 is 7/28, and 1/28 is 1/28, so 7/28 + 1/28 = 8/28 = 2/7. That works. But maybe the problem expects more terms? If the problem allows two terms, then this is correct. But sometimes Egyptian fractions for numbers less than 1 might require more terms. Let me verify.Alternatively, perhaps there's another combination. Let me try another approach. Suppose I start with a smaller unit fraction. Let's try 1/5. 1/5 is 0.2, and 2/7 is approximately 0.2857. So subtract 1/5:2/7 - 1/5 = (10/35 - 7/35) = 3/35.Now, I need to express 3/35 as a sum of unit fractions. The largest unit fraction less than 3/35 is 1/12, since 3/35 ≈ 0.0857, and 1/12 ≈ 0.0833. Subtracting 1/12:3/35 - 1/12 = (36/420 - 35/420) = 1/420.So 2/7 = 1/5 + 1/12 + 1/420. That's three terms. Are all denominators different? 5, 12, 420. Yes. So that's another representation with three unit fractions. But the previous one with two terms also works. So maybe both are valid. But the problem says "several," which could mean at least two. But maybe the simplest is with two terms. However, in Egyptian fractions, sometimes they prefer the minimal number of terms. So 1/4 + 1/28 is minimal here. But maybe the problem expects more. Let me check if there's a way to do it with three terms without getting such a large denominator.Alternatively, maybe starting with 1/4 as the first term is better. Then 1/4 + 1/28. But perhaps there's another way. Let's see.Wait, perhaps starting with 1/5 is not the most efficient. Let's try 1/3. 1/3 is about 0.333, which is larger than 2/7, so that's no good. Then 1/4 is next. As before. Then 1/4 and 1/28. Alternatively, maybe there's a way to split 2/7 into three unit fractions. Let's see.Suppose I try 1/5 + 1/6 + something. Let's compute 1/5 + 1/6 = 11/30. 2/7 is approximately 0.2857, and 11/30 is approximately 0.3667, which is too big. So that's not possible.Alternatively, maybe 1/6 + 1/7. Let's compute that: 1/6 + 1/7 = 13/42 ≈ 0.3095. Still larger than 2/7 (≈0.2857). So subtracting that would not work.Alternatively, try 1/7. 1/7 is about 0.1429. 2/7 - 1/7 = 1/7. But that would give 1/7 + 1/7, but they are the same, which is not allowed. So that's invalid.Alternatively, 1/8. 1/8 = 0.125. 2/7 - 1/8 = (16/56 - 7/56) = 9/56. Now, represent 9/56 as a sum of unit fractions. The largest unit fraction less than 9/56 is 1/7 (since 9/56 ≈0.1607, and 1/7≈0.1429). So subtract 1/7: 9/56 - 1/7 = 9/56 - 8/56 = 1/56. So then 2/7 = 1/8 + 1/7 + 1/56. But wait, 1/7 and 1/56: denominators 8,7,56. All different. So that's three terms. That's another possible expansion.But again, comparing to the previous one, 1/4 +1/28 is simpler. So maybe the problem expects the minimal number of terms, which is two. However, perhaps the problem wants more than two? Let me check the original question again. It says "several different common fractions," and "several" could be two or more. But the example in Egyptian fractions for 2/7 is indeed 1/4 + 1/28, so that's the standard decomposition. So maybe that's the answer expected here.But let me verify once more. Let's check another method. Suppose we use the identity that 2/n = 1/(n+1)/2 + 1/(n(n+1)/2). But this might not always work. Let's see. For n=7: 2/7 = 1/((7+1)/2) + 1/(7*(7+1)/2). (7+1)/2 = 4, and 7*8/2 = 28. So 2/7 = 1/4 + 1/28. Yep, exactly the decomposition we had earlier. So that's a formulaic way to get it. So that seems to be the standard.Alternatively, if the problem requires more than two terms, perhaps we can split 1/28 further. Let's try that. 1/28 can be expressed as another sum of unit fractions. Let's see. The largest unit fraction less than 1/28 is 1/29. But 1/29 is approximately 0.0345, and 1/28 is approximately 0.0357. Then subtract 1/29: 1/28 -1/29 = (29 -28)/(28*29) = 1/812. So 1/28 = 1/29 +1/812. Therefore, 2/7 =1/4 +1/29 +1/812. Now, that's three terms, but they are all distinct. However, this seems unnecessarily complicated, and the denominators are getting very large. Since the problem doesn't specify a preference for the number of terms or the size of denominators, but just requires them to be different, either decomposition is valid. But the most straightforward and standard one is 1/4 +1/28.Therefore, unless the problem specifically requires more than two terms, the answer is 1/4 +1/28. Let me check the problem statement again: "Represent the number 2/7 as the sum of several different common fractions, the numerators of which are equal to one." The term "several" can mean more than one, so two is acceptable. However, sometimes people use "several" to mean more than two, but I think in mathematical contexts, two can be considered several. But to be safe, maybe check if there's a three-term decomposition without using very large denominators. Earlier, we had 1/8 +1/7 +1/56. Let's check that sum: 1/8 +1/7 = 7/56 +8/56 =15/56. Then adding 1/56 gives 16/56 = 4/14 = 2/7. So that works. So that's another valid decomposition with three terms.Alternatively, another decomposition: 1/5 +1/12 +1/420. Let's check that. 1/5 is 84/420, 1/12 is 35/420, and 1/420 is 1/420. Summing those: 84 +35 +1=120. 120/420 = 2/7. Correct. So that's another three-term decomposition. So there are multiple ways. But which one is the correct answer?The problem doesn't specify any constraints other than the fractions being different and having numerator one. So technically, any valid decomposition is acceptable. However, typically, the minimal number of terms is preferred, which would be two: 1/4 +1/28.Alternatively, the problem might expect the most "natural" decomposition, which in Egyptian fractions is usually the greedy algorithm result, which is 1/4 +1/28. So probably that's the answer expected here.But let's confirm with another approach. Let me think about other possible combinations. For example, 1/6 + something. Let's see. 2/7 -1/6 = (12/42 -7/42)=5/42. Now, 5/42. The largest unit fraction less than 5/42 is 1/9 (approx 0.111, 5/42≈0.119). So 5/42 -1/9 = (15/126 -14/126)=1/126. So 2/7=1/6 +1/9 +1/126. Check the sum: 1/6=21/126, 1/9=14/126, 1/126=1/126. Total is 21+14+1=36. 36/126=6/21=2/7. Correct. So that's another three-term decomposition. So there are multiple solutions.But again, the simplest is two terms. Since the problem didn't specify a minimum or maximum number of terms, just "several", which could mean two or more. So the answer is not unique. But perhaps the expected answer is the two-term one. However, to ensure that I cover all possibilities, maybe I should present the two-term solution as the answer, since it's the minimal and standard one.Alternatively, maybe the problem expects three terms. Let me check historical Egyptian fractions. Looking it up mentally, the ancient Egyptian representation for 2/7 is indeed 1/4 + 1/28, so that's the standard one. Therefore, that's the answer they are likely expecting.Therefore, the answer is 1/4 + 1/28.Final AnswerThe number (frac{2}{7}) can be represented as the sum of different unit fractions: boxed{frac{1}{4} + frac{1}{28}}.</think>To represent the number (frac{2}{7}) as the sum of several different common fractions with numerators equal to one, we can use the greedy algorithm, which involves taking the largest possible unit fraction less than the remaining amount and repeating the process.1. Start with (frac{2}{7}). The largest unit fraction less than (frac{2}{7}) is (frac{1}{4}).2. Subtract (frac{1}{4}) from (frac{2}{7}): [ frac{2}{7} - frac{1}{4} = frac{8}{28} - frac{7}{28} = frac{1}{28} ]3. This results in (frac{2}{7} = frac{1}{4} + frac{1}{28}).Both fractions (frac{1}{4}) and (frac{1}{28}) are different unit fractions. This decomposition uses the minimal number of terms and is a standard Egyptian fraction representation.Thus, the number (frac{2}{7}) can be represented as the sum of different unit fractions: boxed{frac{1}{4} + frac{1}{28}}.]

🔑:# SolutionWe use the fact thatfrac{1}{n}=frac{1}{n+1}+frac{1}{n(n+1)}Thenfrac{2}{7}=frac{1}{7}+frac{1}{7}=frac{1}{7}+frac{1}{8}+frac{1}{56}Other variations are also possible.Any other correct representation was also accepted as an answer.

🧠:[<think>Okay, so I need to find positive primes p and q such that the equation x⁴ + p²x + q = 0 has a multiple root. Hmm, let's see. I remember that a polynomial equation has a multiple root if and only if the polynomial and its derivative share a common root. So, first, I should find the derivative of the given polynomial and then set up the conditions for a common root.The polynomial is f(x) = x⁴ + p²x + q. Its derivative f'(x) would be 4x³ + p². So, if α is a multiple root, then both f(α) = 0 and f'(α) = 0 must hold. Let me write that down:1. α⁴ + p²α + q = 02. 4α³ + p² = 0From equation 2, I can solve for p² in terms of α: p² = -4α³. But p is a prime, so p² is positive. However, -4α³ is negative unless α is a negative real number. So α must be negative. Let me denote α = -β where β > 0. Then substituting into equation 2: p² = -4(-β)³ = -4*(-β³) = 4β³. So p² = 4β³. Since p is prime, p² is a square of a prime, which must be 4β³. Therefore, 4β³ must be a square of a prime. Let's write 4β³ = p². Then β³ = p² / 4. Since β is a real number, here we can consider β as a positive rational number because p is a prime (integer) and we need β³ to be a rational number. Hmm, maybe β is a fraction?Alternatively, maybe β is an integer? Let me check. If β is an integer, then β³ must be equal to (p²)/4. But p² is a square of a prime, so p² is 4, 9, 25, 49, etc. So (p²)/4 must be an integer if β is an integer. Therefore, p² must be divisible by 4, which implies that p must be even. The only even prime is 2. Therefore, if p = 2, then p² = 4, so β³ = 4/4 = 1, so β = 1. Then α = -β = -1. Let me check if this works.If p = 2, then α = -1. Then plug into equation 1: (-1)^4 + (2)^2*(-1) + q = 0. That is, 1 + 4*(-1) + q = 0 => 1 - 4 + q = 0 => q = 3. Since q is a prime, 3 is prime. So (p, q) = (2, 3) is a possible solution.But wait, maybe there are other primes where β is not an integer. Let's see. Suppose β is a fraction, say β = a/b in reduced form where a and b are positive integers. Then β³ = a³ / b³ = p² / 4. Therefore, 4a³ = p² b³. So p² divides 4a³. Since p is a prime, p divides 4a³. If p divides 4, then p = 2. If p divides a³, then p divides a. So if p ≠ 2, then p divides a. Let's consider two cases: p = 2 and p odd prime.Case 1: p = 2. Then p² = 4. Then 4β³ = 4 => β³ = 1 => β = 1, so α = -1. Then as above, q = 3. So that's one solution.Case 2: p is an odd prime. Then p divides a, so let a = p*k where k is a positive integer. Then substituting into 4a³ = p² b³: 4*(p³ k³) = p² b³ => 4p³ k³ = p² b³ => 4p k³ = b³. So b³ is divisible by p. Since p is a prime and we assumed β = a/b is in reduced form, but a = p*k and b must now have a factor of p, which would contradict the reduced form. Wait, unless k and b have common factors. Wait, maybe I need to re-examine this.Wait, if β = a/b is in reduced form, then a and b are coprime. But we have a = p*k, so a and b are coprime implies that p does not divide b. However, from 4p k³ = b³, since p is an odd prime dividing the left-hand side, and p does not divide b (since a and b are coprime and p divides a), then p must divide 4p k³, which is true, but on the right-hand side, b³, since p does not divide b, p does not divide b³. Therefore, the equation 4p k³ = b³ implies that p divides the left-hand side but not the right-hand side, which is a contradiction unless p divides 4. But p is an odd prime, so p doesn't divide 4. Hence, the only possibility is that this case leads to a contradiction, meaning that there are no solutions when p is an odd prime.Therefore, the only possible solution is p = 2, q = 3.Wait, but let me verify this conclusion again. Because maybe even though there is a contradiction in the case when p is an odd prime, perhaps there's another approach. Let me check.Suppose p is an odd prime. Then from equation 2: p² = -4α³. Since α is negative, let α = -β as before. Then p² = 4β³. So p² must be divisible by 4, hence p must be even. But the only even prime is 2, so p must be 2. Therefore, there are no solutions with p being an odd prime. Hence, p must be 2, leading to q = 3. Therefore, the only solution is p = 2 and q = 3.But let me check this by plugging back into the original equation. If p = 2 and q = 3, then the equation becomes x⁴ + 4x + 3 = 0. Let's check if it has a multiple root. We found that α = -1 is a root. Let's see if it's a multiple root. Compute f(-1) = (-1)^4 + 4*(-1) + 3 = 1 - 4 + 3 = 0. Then f'(-1) = 4*(-1)^3 + 4 = -4 + 4 = 0. So yes, x = -1 is a multiple root. Are there other multiple roots?Suppose there is another multiple root, say γ. Then f(γ) = 0 and f'(γ) = 0. So similar equations:γ⁴ + 4γ + 3 = 04γ³ + 4 = 0 => γ³ = -1 => γ = -1. So the only multiple root is γ = -1. Therefore, the equation x⁴ + 4x + 3 has a single multiple root at x = -1. Therefore, this is the only solution.But let me check if there could be other primes p and q where perhaps there's a different multiple root. Suppose we don't assume α is real, but complex. But since the coefficients are real, any complex roots would come in conjugate pairs. However, multiple roots would still have to satisfy f(α) = 0 and f'(α) = 0. But the equations would still require p² = -4α³, so α³ must be negative real, hence α is a negative real number. So even if there are complex roots, the multiple root must be real and negative. Therefore, the previous analysis holds.Therefore, the only solution is p = 2 and q = 3. Let me confirm once again. Suppose p = 2, q = 3, then the equation is x⁴ + 4x + 3 = 0. Let's factor it. Since x = -1 is a double root, we can factor (x + 1)^2 out. Let's perform polynomial division.Divide x⁴ + 4x + 3 by (x + 1)^2 = x² + 2x + 1.Using polynomial long division:Divide x⁴ by x²: x². Multiply (x² + 2x + 1) by x²: x⁴ + 2x³ + x². Subtract from x⁴ + 0x³ + 0x² + 4x + 3: gives -2x³ - x² + 4x + 3.Next, divide -2x³ by x²: -2x. Multiply (x² + 2x + 1) by -2x: -2x³ -4x² -2x. Subtract: (-2x³ - x² + 4x + 3) - (-2x³ -4x² -2x) = 3x² + 6x + 3.Next, divide 3x² by x²: 3. Multiply (x² + 2x + 1) by 3: 3x² + 6x + 3. Subtract: 0. So the division yields x² - 2x + 3. Therefore, x⁴ + 4x + 3 = (x + 1)^2 (x² - 2x + 3). The quadratic factor x² - 2x + 3 has discriminant 4 - 12 = -8, so it has complex roots. Therefore, the equation indeed has a double root at x = -1 and two complex roots. Therefore, this checks out.Is there any other possibility? Let's see. Suppose that p and q are primes such that there exists a multiple root. The above analysis suggests that p must be 2, leading to q = 3. Let me check for p = 2, q = 3. Are these primes? Yes, 2 and 3 are primes. So that's the solution.Therefore, the answer is p = 2 and q = 3.Final AnswerThe positive primes are boxed{2} and boxed{3}.</think>To determine the positive primes ( p ) and ( q ) for which the equation ( x^4 + p^2 x + q = 0 ) has a multiple root, we need to check when the polynomial and its derivative share a common root. Given the polynomial ( f(x) = x^4 + p^2 x + q ), its derivative is ( f'(x) = 4x^3 + p^2 ). If ( alpha ) is a multiple root, then both ( f(alpha) = 0 ) and ( f'(alpha) = 0 ) must hold:1. ( alpha^4 + p^2 alpha + q = 0 )2. ( 4alpha^3 + p^2 = 0 )From the second equation, solving for ( p^2 ) gives ( p^2 = -4alpha^3 ). Since ( p ) is a prime, ( p^2 ) must be positive, implying ( alpha ) is negative. Let ( alpha = -beta ) where ( beta > 0 ). Substituting ( alpha = -beta ) into ( p^2 = -4alpha^3 ) gives ( p^2 = 4beta^3 ). For ( p ) to be a prime, ( 4beta^3 ) must be the square of a prime. The only even prime is 2, so setting ( p = 2 ) gives ( p^2 = 4 ). Thus, ( 4beta^3 = 4 ) implies ( beta^3 = 1 ), so ( beta = 1 ). Substituting ( beta = 1 ) back, we get ( alpha = -1 ).Checking ( alpha = -1 ) in the first equation:[(-1)^4 + 2^2(-1) + q = 0 implies 1 - 4 + q = 0 implies q = 3]Verification shows that ( x = -1 ) is indeed a multiple root of ( x^4 + 4x + 3 = 0 ), confirming that ( p = 2 ) and ( q = 3 ) are primes. Thus, the positive primes ( p ) and ( q ) are (boxed{2}) and (boxed{3}).,<think>Okay, so I need to find positive prime numbers p and q such that the equation x⁴ + p²x + q = 0 has a multiple root. Hmm, multiple roots... I remember from algebra that if a polynomial has a multiple root, then that root is also a root of its derivative. So maybe I should start by taking the derivative of the given polynomial and then find common roots between the original polynomial and its derivative?Let me write down the polynomial and its derivative. The polynomial is f(x) = x⁴ + p²x + q. The derivative f'(x) would be 4x³ + p². If there's a multiple root, say α, then α must satisfy both f(α) = 0 and f'(α) = 0. So, we have two equations:1. α⁴ + p²α + q = 02. 4α³ + p² = 0I need to solve these equations simultaneously for α, p, and q, with p and q being primes. Let me see. From equation 2, I can solve for p²: p² = -4α³. But p is a positive prime, so p² is positive. However, the right-hand side is -4α³. For this to hold, α must be negative because then -4α³ would be positive (since α³ would be negative and multiplying by -4 gives positive). So α is a negative real number.Let me denote α as -β where β is positive. Then substituting α = -β into equation 2: 4(-β)³ + p² = 0 → -4β³ + p² = 0 → p² = 4β³. Therefore, p² must be equal to 4β³. Since p is prime, p² has only a few factors: 1, p, and p². But 4β³ = p², so 4 must divide p². Since p is prime, p must be 2 because 2 is the only prime that divides 4. So p = 2. Then p² = 4, so 4 = 4β³ → β³ = 1 → β = 1. Therefore, α = -β = -1.Now, let's check if α = -1 is a root of the original polynomial. Substitute x = -1 into f(x):f(-1) = (-1)⁴ + p²(-1) + q = 1 - p² + q. Since p = 2, this becomes 1 - 4 + q = q - 3. For this to equal 0, q must be 3. Since 3 is a prime number, that works. So p = 2 and q = 3. Are there any other possibilities?Wait, let me double-check. We assumed that p is prime, so p² must be 4β³. If p is another prime, say 3, then p² = 9. Then 9 = 4β³ → β³ = 9/4, which is not an integer. Since β must be a positive real number, but since we are dealing with algebraic equations, β might not necessarily be rational. However, in this case, if β³ = 9/4, β is irrational, but when we substitute back into the original equation, would that still satisfy?Wait, maybe I need to consider that α must be a common root, so perhaps β needs to be rational? Not necessarily. But in our previous steps, we found that when p is 2, β is 1, which is an integer. Let me check if there are other primes p such that 4β³ = p². Since p is prime, p² can only be factored into 1, p, p². So 4β³ must equal p². Therefore, β³ = p² / 4. Since β is real, β = (p² / 4)^(1/3). However, for β to be rational, p² must be a multiple of 4. Since p is prime, the only prime where p² is divisible by 4 is p = 2. Any other prime squared would be like 3² = 9, 5² = 25, etc., which are not divisible by 4. Therefore, the only possible prime p is 2, leading to β³ = 1, so β = 1. Therefore, α = -1. Then substituting back into the original equation gives q = 3.Therefore, the only possible primes are p = 2 and q = 3. Let me verify once again to be sure.If p = 2 and q = 3, the polynomial becomes x⁴ + 4x + 3. Let's check if x = -1 is a multiple root. Compute f(-1): (-1)^4 + 4*(-1) + 3 = 1 - 4 + 3 = 0. Then compute f'(-1): 4*(-1)^3 + 4 = -4 + 4 = 0. So yes, x = -1 is a root of both f and f', hence a multiple root. Therefore, the polynomial has a multiple root at x = -1 when p = 2 and q = 3.Is there any other possible multiple root? Suppose there's another root that's a multiple root. Let me think. Suppose there is another multiple root γ. Then γ would satisfy 4γ³ + p² = 0 and γ⁴ + p²γ + q = 0. But from the derivative equation, p² = -4γ³. Substitute into the original equation: γ⁴ + (-4γ³)γ + q = γ⁴ - 4γ⁴ + q = -3γ⁴ + q = 0 → q = 3γ⁴. But since q is a prime, 3γ⁴ must be prime. The only way this is possible is if γ⁴ = 1 and 3*1 = 3, which is prime. So γ⁴ = 1 implies γ = 1 or γ = -1. But we already considered γ = -1. If γ = 1, then p² = -4*(1)^3 = -4, which is impossible because p² is positive. Therefore, the only possible multiple root is γ = -1, leading to p = 2 and q = 3.Therefore, the primes are p = 2 and q = 3.Final AnswerThe positive primes are boxed{p=2} and boxed{q=3}.</think>To determine the positive primes ( p ) and ( q ) for which the equation ( x^4 + p^2 x + q = 0 ) has a multiple root, we start by noting that a multiple root must satisfy both the polynomial and its derivative. The polynomial is ( f(x) = x^4 + p^2 x + q ) and its derivative is ( f'(x) = 4x^3 + p^2 ). If ( alpha ) is a multiple root, then it must satisfy:1. ( alpha^4 + p^2 alpha + q = 0 )2. ( 4alpha^3 + p^2 = 0 )From the second equation, solving for ( p^2 ) gives:[ p^2 = -4alpha^3 ]Since ( p ) is a positive prime, ( p^2 ) must be positive, implying ( alpha ) must be negative. Let ( alpha = -beta ) where ( beta ) is positive. Substituting ( alpha = -beta ) into the second equation:[ 4(-beta)^3 + p^2 = 0 implies -4beta^3 + p^2 = 0 implies p^2 = 4beta^3 ]Since ( p ) is a prime, ( p^2 ) must be a square of a prime. The equation ( p^2 = 4beta^3 ) implies ( 4 ) must divide ( p^2 ). The only prime ( p ) such that ( p^2 ) is divisible by 4 is ( p = 2 ). Substituting ( p = 2 ):[ 4 = 4beta^3 implies beta^3 = 1 implies beta = 1 ]Thus, ( alpha = -1 ).Substituting ( alpha = -1 ) into the original polynomial:[ (-1)^4 + 2^2(-1) + q = 0 implies 1 - 4 + q = 0 implies q = 3 ]Verifying, the polynomial ( x^4 + 4x + 3 ) has a multiple root at ( x = -1 ) since both ( f(-1) = 0 ) and ( f'(-1) = 0 ). Checking other primes for ( p ) leads to non-integer or non-prime values for ( q ).Thus, the positive primes are (boxed{p=2}) and (boxed{q=3}).]

🔑:Let the multiple (at least double) root of the equation be denoted by a (which must necessarily be non-positive). Then[begin{gathered}x^{4}+p^{2} x+q=(x+a)^{2}left(x^{2}+b x+cright)=left(x^{2}+2 a x+a^{2}right)left(x^{2}+b x+cright)= =x^{4}+(2 a+b) x^{3}+left(2 a b+a^{2}+cright) x^{2}+a(a b+2 c) x+a^{2} cend{gathered}]By comparing the corresponding coefficients, we get(1) 2 a+b=0, hence b=-2 a,(2) 2 a b+a^{2}+c=0, thus c=3 a^{2}(3) a(a b+2 c)=p^{2}, so 4 a^{3}=p^{2},(4) a^{2} c=q, hence 3 a^{4}=q.From (3) and (4),[3^{3} p^{8}=3^{3}left(p^{2}right)^{4}=3^{3} cdot 4^{4} cdot a^{12}=2^{8}left(3 a^{4}right)^{3}=2^{8} q^{3}]This is only possible if p=2 and q=3. Then a=1, b=-2, c=3, and indeed x^{4}+4 x+3=(x+1)^{2}left(x^{2}-2 x+3right). Gergely Ambrus (Szeged, Radnóti M. Gymnasium, 12th grade)

🧠:[<think>Okay, let's see. I need to find the minimum value of the function y = 4/(x - 2) + x when x is greater than 2. The options are A:5, B:4, C:6, D:8. Hmm, alright. Let me think about how to approach this. First, since we're dealing with a function that has a variable in both the denominator and the numerator, maybe calculus would help here. If I can find the derivative of the function and set it equal to zero, that should give me the critical points where the function might have a minimum or maximum. Then I can check if that point is a minimum by using the second derivative test or analyzing the behavior around that point. So, let's start by finding the derivative of y with respect to x. The function is y = 4/(x - 2) + x. Let's write that as y = 4*(x - 2)^(-1) + x. Taking the derivative term by term. The derivative of 4*(x - 2)^(-1) is -4*(x - 2)^(-2)*1, using the chain rule. The derivative of x is 1. So putting that together, dy/dx = -4/(x - 2)^2 + 1. Now, to find the critical points, set dy/dx = 0:-4/(x - 2)^2 + 1 = 0Let me solve for x. Let's move the -4/(x - 2)^2 to the other side:1 = 4/(x - 2)^2Multiply both sides by (x - 2)^2:(x - 2)^2 = 4Take the square root of both sides:x - 2 = ±2But since x > 2 (given in the problem), x - 2 must be positive. So we discard the negative solution:x - 2 = 2Therefore, x = 4. So the critical point is at x = 4. Now we need to check if this is a minimum. Let's do the second derivative test. First, compute the second derivative. The first derivative was dy/dx = -4/(x - 2)^2 + 1. So the second derivative d²y/dx² is the derivative of that. The derivative of -4/(x - 2)^2 is (-4)*(-2)*(x - 2)^(-3) = 8/(x - 2)^3. The derivative of 1 is 0. So the second derivative is 8/(x - 2)^3. Now evaluate the second derivative at x = 4. d²y/dx² = 8/(4 - 2)^3 = 8/(2)^3 = 8/8 = 1. Since the second derivative is positive (1 > 0) at x = 4, this means the function is concave upward there, so the critical point at x = 4 is indeed a local minimum. Since the function approaches infinity as x approaches 2 from the right (because the term 4/(x - 2) becomes very large) and as x approaches infinity (because the x term dominates and grows without bound), the local minimum at x = 4 should be the global minimum on the interval x > 2. Therefore, substitute x = 4 into the original function to find the minimum value:y = 4/(4 - 2) + 4 = 4/2 + 4 = 2 + 4 = 6. Wait, that gives 6, which is option C. But let me double-check my calculations to make sure I didn't make a mistake. Original function: y = 4/(x - 2) + x. At x = 4: 4/(4 - 2) = 4/2 = 2. Then 2 + 4 = 6. That seems right. But wait, let me check the answer options again. The options are A:5, B:4, C:6, D:8. So according to this, the answer is C:6. But I was expecting maybe 4? Hmm. Wait, maybe I did a mistake in the derivative? Let me check again. First derivative: dy/dx = -4/(x - 2)^2 + 1. Correct. Setting that to zero: -4/(x - 2)^2 + 1 = 0. Then (x - 2)^2 = 4. So x - 2 = 2, so x = 4. Correct. Second derivative: 8/(x - 2)^3. At x =4, that's 8/8 =1, which is positive. So minimum. Then y at x=4 is 6. So answer is C:6. But wait, maybe the problem is written incorrectly? Let me check again the original problem. It says: Given x > 2, the minimum value of the function y = 4/(x - 2) + x is ( ). The options are A:5, B:4, C:6, D:8. So according to my calculations, the answer is C:6. But let me think again. Maybe there's another approach. Sometimes, for functions of the form y = a/(x - b) + x, we can use the AM-GM inequality. Let's see if that applies here. The AM-GM inequality states that for positive real numbers, the arithmetic mean is greater than or equal to the geometric mean. So if I can write the expression as a sum of two positive terms, perhaps I can apply AM-GM. Let me rewrite the function. Let's denote t = x - 2. Since x > 2, then t > 0. Then x = t + 2, and substituting into y gives:y = 4/t + t + 2. So y = t + 4/t + 2. Now, we can consider the terms t and 4/t. Since t > 0, both terms are positive. Applying AM-GM to t and 4/t:(t + 4/t)/2 ≥ √(t * 4/t) = √4 = 2. Multiplying both sides by 2:t + 4/t ≥ 4. Therefore, y = t + 4/t + 2 ≥ 4 + 2 = 6. Equality holds when t = 4/t, which implies t^2 = 4, so t = 2 (since t > 0). Then t = 2 implies x - 2 = 2, so x = 4. Therefore, the minimum value of y is 6, achieved at x =4. So this also gives answer C:6. Therefore, both calculus and AM-GM methods lead to the same conclusion. So the correct answer should be C:6. But let me check once again. Maybe there's a trick here. Let me plot the function or test some values. For example, if x=3: y=4/(1)+3=4+3=7. Which is higher than 6.x=4: y=4/2 +4=2+4=6. x=5: y=4/3 +5≈1.33+5=6.33, which is higher.x=6: y=4/4 +6=1 +6=7. x approaches 2 from the right: 4/(very small) + 2 approaches infinity.x approaches infinity: x dominates, so y approaches infinity.So indeed, the minimum occurs at x=4 with y=6. So answer is C:6. But wait, the options were A:5, B:4, C:6, D:8. So C is the correct answer. But maybe I messed up in the AM-GM approach. Let me check that again. Original substitution: t = x -2, so y = 4/t + t +2. Then applying AM-GM on t and 4/t. But AM-GM says (a + b)/2 ≥ √(ab). So here a = t, b =4/t. Then (t + 4/t)/2 ≥ √(t *4/t) = √4 =2. So t +4/t ≥4. Then adding 2 gives y ≥6. So equality when t=4/t, t=2. Yes, that's correct. So all steps are valid. Therefore, answer is C:6. So why is there an option B:4? Maybe if someone forgets to add the 2 in the substitution? Let's see. If someone did AM-GM on 4/(x-2) and x, that would be incorrect because x is not a positive term independent of x-2. Wait, x can be written as (x -2) + 2. So perhaps trying to apply AM-GM directly on 4/(x -2) and (x -2) + 2. But that's more complicated. Alternatively, suppose someone differentiates incorrectly. For example, if someone thought the derivative of 4/(x -2) is -4/(x -2)^2, which is correct, and adds derivative of x, which is 1. Then sets to zero. So the same result. So I think my answer is correct. Therefore, the answer is C:6. Final Answerboxed{C}</think>To find the minimum value of the function ( y = dfrac{4}{x - 2} + x ) for ( x > 2 ), we can use calculus and the AM-GM inequality.Calculus Approach:1. First Derivative: [ frac{dy}{dx} = -frac{4}{(x - 2)^2} + 1 ] Setting the derivative equal to zero to find critical points: [ -frac{4}{(x - 2)^2} + 1 = 0 implies frac{4}{(x - 2)^2} = 1 implies (x - 2)^2 = 4 implies x - 2 = 2 implies x = 4 ] (Since ( x > 2 ), we discard the negative solution.)2. Second Derivative Test: [ frac{d^2y}{dx^2} = frac{8}{(x - 2)^3} ] Evaluating at ( x = 4 ): [ frac{d^2y}{dx^2} = frac{8}{(4 - 2)^3} = frac{8}{8} = 1 > 0 ] Since the second derivative is positive, ( x = 4 ) is a local minimum.3. Evaluating the Function at ( x = 4 ): [ y = frac{4}{4 - 2} + 4 = frac{4}{2} + 4 = 2 + 4 = 6 ]AM-GM Inequality Approach:1. Substitution: Let ( t = x - 2 ) (where ( t > 0 )), then ( x = t + 2 ). The function becomes: [ y = frac{4}{t} + t + 2 ]2. Applying AM-GM: [ frac{t + frac{4}{t}}{2} geq sqrt{t cdot frac{4}{t}} implies frac{t + frac{4}{t}}{2} geq 2 implies t + frac{4}{t} geq 4 ] Adding 2 to both sides: [ y = t + frac{4}{t} + 2 geq 4 + 2 = 6 ] Equality holds when ( t = 2 ), which implies ( x = 4 ).Both methods confirm that the minimum value of the function is ( 6 ), achieved at ( x = 4 ).[boxed{C}],<think>Okay, so I need to find the minimum value of the function y = 4/(x - 2) + x when x is greater than 2. The options are A:5, B:4, C:6, D:8. Let me think about how to approach this.First, I remember that to find the minimum or maximum of a function, calculus is usually involved, like finding the derivative and setting it to zero. Since this function has a variable in the denominator and a linear term, taking the derivative might work. Alternatively, maybe there's a way to use algebra, like completing the square or using the AM-GM inequality. Hmm, let me try both approaches and see which one makes sense.Starting with calculus. The function is y = 4/(x - 2) + x. Let's find its derivative to locate critical points.The derivative of 4/(x - 2) with respect to x is -4/(x - 2)^2, using the power rule. The derivative of x with respect to x is 1. So, the derivative y' = -4/(x - 2)^2 + 1.To find critical points, set y' = 0:-4/(x - 2)^2 + 1 = 0Let's solve for x. Moving the second term to the other side:-4/(x - 2)^2 = -1Multiply both sides by -1:4/(x - 2)^2 = 1Multiply both sides by (x - 2)^2:4 = (x - 2)^2Take square roots of both sides:sqrt(4) = |x - 2|So, 2 = |x - 2|Since x > 2, the absolute value |x - 2| is just x - 2, so:2 = x - 2Therefore, x = 4.So, the critical point is at x = 4. Now, we need to check if this is a minimum. Since it's the only critical point for x > 2, we can test the second derivative or check values around x = 4.Let me try the second derivative. First derivative is y' = -4/(x - 2)^2 + 1. Second derivative y'':The derivative of -4/(x - 2)^2 is -4 * (-2)/(x - 2)^3 = 8/(x - 2)^3. The derivative of 1 is 0. So, y'' = 8/(x - 2)^3.At x = 4, y'' = 8/(4 - 2)^3 = 8/(8) = 1. Since y'' is positive at x = 4, this means the function is concave upwards there, so it's a local minimum. Since there's only one critical point in the domain x > 2, this must be the global minimum.Therefore, plugging x = 4 back into the original function:y = 4/(4 - 2) + 4 = 4/2 + 4 = 2 + 4 = 6.So, the minimum value is 6, which is option C. Let me check if I made any mistakes here.Alternatively, maybe using AM-GM inequality. Let's see. The function is y = 4/(x - 2) + x. Let me make a substitution to simplify. Let t = x - 2, since x > 2, so t > 0. Then, x = t + 2. Substitute into y:y = 4/t + (t + 2) = 4/t + t + 2.Now, we can consider the expression 4/t + t. The AM-GM inequality states that for positive real numbers a and b, the arithmetic mean is greater than or equal to the geometric mean: (a + b)/2 ≥ sqrt(ab). Equality holds when a = b.So, applying AM-GM to 4/t and t:(4/t + t)/2 ≥ sqrt(4/t * t) = sqrt(4) = 2Multiply both sides by 2:4/t + t ≥ 4Therefore, y = 4/t + t + 2 ≥ 4 + 2 = 6. Equality holds when 4/t = t, so t^2 = 4, t = 2 (since t > 0). Therefore, t = 2, which means x = t + 2 = 4. So, the minimum value is 6, achieved at x = 4. That matches the calculus result.Therefore, the answer should be C:6.But let me double-check with some test values. Let's pick x = 3. Then y = 4/(3 - 2) + 3 = 4 + 3 = 7, which is more than 6. x = 5: y = 4/3 + 5 ≈ 1.333 + 5 = 6.333, which is more than 6. x = 4: y = 6. If I go closer to 2, like x = 2.5, then y = 4/(0.5) + 2.5 = 8 + 2.5 = 10.5, which is much larger. So, yes, 6 is the minimum. The answer is C.Final Answerboxed{C}</think>To find the minimum value of the function ( y = dfrac{4}{x - 2} + x ) for ( x > 2 ), we can use calculus and the AM-GM inequality.Calculus Approach:1. Find the derivative of the function: [ y' = -dfrac{4}{(x - 2)^2} + 1 ]2. Set the derivative to zero to find critical points: [ -dfrac{4}{(x - 2)^2} + 1 = 0 implies dfrac{4}{(x - 2)^2} = 1 implies (x - 2)^2 = 4 implies x - 2 = 2 implies x = 4 ]3. Check the second derivative to confirm it's a minimum: [ y'' = dfrac{8}{(x - 2)^3} ] At ( x = 4 ): [ y'' = dfrac{8}{(4 - 2)^3} = 1 > 0 implies text{Concave upwards, hence a minimum} ]4. Calculate the minimum value: [ y = dfrac{4}{4 - 2} + 4 = 2 + 4 = 6 ]AM-GM Inequality Approach:1. Substitute ( t = x - 2 ) (where ( t > 0 )): [ y = dfrac{4}{t} + t + 2 ]2. Apply AM-GM inequality to ( dfrac{4}{t} ) and ( t ): [ dfrac{dfrac{4}{t} + t}{2} geq sqrt{dfrac{4}{t} cdot t} = 2 implies dfrac{4}{t} + t geq 4 ] [ y geq 4 + 2 = 6 ] Equality holds when ( dfrac{4}{t} = t implies t = 2 implies x = 4 ).Both methods confirm that the minimum value is 6, achieved at ( x = 4 ).[boxed{C}]]

🔑:Analysis: Since (x > 2), we have (x - 2 > 0). Thus, (y = dfrac{4}{x - 2} + x = dfrac{4}{x - 2} + (x - 2) + 2). We can apply the basic inequality to find the minimum value. Given (x > 2), so (x - 2 > 0), then (y = dfrac{4}{x - 2} + x = dfrac{4}{x - 2} + (x - 2) + 2) (geqslant 2 sqrt{ dfrac{4}{x - 2} cdot (x - 2)} + 2 = 6), equality holds if and only if (dfrac{4}{x - 2} = x - 2), i.e., (x = 4), thus, the minimum value of the function (y = dfrac{4}{x - 2} + x) is (6). Therefore, the correct answer is boxed{C}.

🧠:[<think>Okay, so I need to factorize the expression ( x^2 + 2xy + y^2 ). Hmm, let's see. I remember from algebra that factoring quadratic expressions often involves looking for patterns or special products. Let me think... Oh, right! There are some common formulas for factoring, like the difference of squares, perfect square trinomials, etc. Wait, this expression has three terms: ( x^2 ), ( 2xy ), and ( y^2 ). The first thing I notice is that the first and last terms are both squares. ( x^2 ) is obviously ( x ) squared, and ( y^2 ) is ( y ) squared. The middle term is ( 2xy ), which is twice the product of ( x ) and ( y ). Hold on, that reminds me of the formula for a perfect square trinomial. The formula is ( (a + b)^2 = a^2 + 2ab + b^2 ). Let me check if this matches the given expression. If I let ( a = x ) and ( b = y ), then substituting into the formula gives:( (x + y)^2 = x^2 + 2xy + y^2 ).Which is exactly the same as the given expression. So, does that mean ( x^2 + 2xy + y^2 ) factors into ( (x + y)^2 )? Let me verify by expanding ( (x + y)^2 ) to make sure.Expanding ( (x + y)^2 ):First, multiply ( (x + y)(x + y) ).Using the distributive property: ( x(x + y) + y(x + y) ).Which is ( x^2 + xy + xy + y^2 ).Combining like terms: ( x^2 + 2xy + y^2 ).Yes, that's exactly the original expression. So the factorization is correct. Therefore, the expression ( x^2 + 2xy + y^2 ) factors to ( (x + y)^2 ).Just to be thorough, let me consider if there's another way this could be factored. For example, sometimes trinomials can be factored into two binomials even if they aren't perfect squares. Let's check if that's possible here.The standard form of a quadratic trinomial is ( ax^2 + bx + c ). In this case, if we treat ( x ) as the variable and ( y ) as a constant, the expression is ( x^2 + 2xy + y^2 ). So, comparing to ( ax^2 + bx + c ), ( a = 1 ), ( b = 2y ), and ( c = y^2 ).To factor it, we look for two numbers that multiply to ( a times c = 1 times y^2 = y^2 ) and add up to ( b = 2y ). The numbers would have to be ( y ) and ( y ), because ( y times y = y^2 ) and ( y + y = 2y ). Therefore, splitting the middle term: ( x^2 + xy + xy + y^2 ). Then grouping:( (x^2 + xy) + (xy + y^2) )Factor out the common terms in each group:( x(x + y) + y(x + y) )Now factor out the common binomial ( (x + y) ):( (x + y)(x + y) = (x + y)^2 )So even when using the standard factoring method for trinomials, we end up with the same result. This confirms that the factorization is indeed ( (x + y)^2 ).I think that's all. The key was recognizing the structure of a perfect square trinomial. Since both the first and last terms are squares and the middle term is twice the product of their square roots, it fits the pattern perfectly. No need to overcomplicate it. Just apply the formula and verify by expanding. Yep, seems solid.Final AnswerThe factorized form of ( x^2 + 2xy + y^2 ) is boxed{(x + y)^2}.</think>To factorize the expression ( x^2 + 2xy + y^2 ), we recognize it as a perfect square trinomial. The formula for a perfect square trinomial is ( (a + b)^2 = a^2 + 2ab + b^2 ). Here, we let ( a = x ) and ( b = y ). Substituting these into the formula, we get:[(x + y)^2 = x^2 + 2xy + y^2]We verify this by expanding ( (x + y)^2 ):[(x + y)(x + y) = x(x + y) + y(x + y) = x^2 + xy + xy + y^2 = x^2 + 2xy + y^2]This matches the original expression. Additionally, using the standard factoring method for trinomials, we split the middle term and factor by grouping:[x^2 + 2xy + y^2 = x^2 + xy + xy + y^2 = (x^2 + xy) + (xy + y^2) = x(x + y) + y(x + y) = (x + y)(x + y) = (x + y)^2]Thus, the factorized form of ( x^2 + 2xy + y^2 ) is boxed{(x + y)^2}.,<think>Okay, so I need to factorize the expression x squared plus 2xy plus y squared. Hmm, let me think. Alright, first, I remember that factoring quadratic expressions often involves looking for patterns or common factors. The given expression is x² + 2xy + y². Let me check if there's a common factor in all the terms. Well, x and y are variables, and the coefficients are 1, 2, and 1. There's no common numerical factor, so that's probably not the way to go.Wait, maybe this is a perfect square trinomial. I recall that a perfect square trinomial has the form (a + b)², which expands to a² + 2ab + b². Comparing that to the given expression, let's see:(a + b)² = a² + 2ab + b²So if I let a = x and b = y, then substituting into the formula gives:(x + y)² = x² + 2xy + y²Which matches exactly the expression we need to factorize. Therefore, the factorization should be (x + y) squared. Let me double-check that by expanding (x + y)² to ensure it gives back the original expression.Expanding (x + y)²:First, multiply x and y: (x + y)(x + y)Using the distributive property:x(x + y) + y(x + y) = x² + xy + xy + y² = x² + 2xy + y²Yep, that's exactly the same as the original expression. So the factorization is correct. I think that's all there is to it. It was a straightforward application of the perfect square trinomial formula. I don't see any other factors here because the expression is a binomial square, so it can't be factored further using real numbers. Let me just confirm once more that there aren't any other methods or factors I might be missing.Alternatively, sometimes people factor by grouping, but in this case, since there are three terms, grouping would be splitting into groups. Let me try that method just to verify. So, x² + 2xy + y². If I try to split the middle term, but since the coefficient is 2, splitting might not be necessary. Let's see:x² + 2xy + y². If I consider the first two terms: x² + 2xy, which is x(x + 2y), but then the last term is y². That doesn't seem helpful. Alternatively, maybe grouping as (x² + y²) + 2xy. But x² + y² is itself a sum of squares, which doesn't factor over the reals, and adding 2xy to it. Hmm, not sure. That approach doesn't seem to lead anywhere. So the perfect square approach is definitely the right way here.Another check: If I substitute some numbers into the original expression and the factored form to see if they match. Let's take x = 1 and y = 1.Original expression: 1² + 2*1*1 + 1² = 1 + 2 + 1 = 4Factored form: (1 + 1)² = 2² = 4. They match. Let's try another set, like x = 2, y = 3.Original: 2² + 2*2*3 + 3² = 4 + 12 + 9 = 25Factored: (2 + 3)² = 5² = 25. Also matches. So that confirms that the factorization works numerically as well.Therefore, I'm confident that the correct factorization of x² + 2xy + y² is (x + y)². There's no need for further factoring here because it's already a perfect square, and in the realm of real numbers, we can't factor it any more. If we were dealing with complex numbers, but the question doesn't specify that, so I think real numbers are assumed here.In summary, recognizing the structure of a perfect square trinomial made this problem straightforward. The key was noticing that the first and last terms are squares of x and y, and the middle term is twice the product of x and y. That's the classic sign of a perfect square, so applying the formula gives the solution directly.Final AnswerThe factorized form is boxed{(x + y)^2}.</think>To factorize the expression (x^2 + 2xy + y^2), we recognize that it is a perfect square trinomial. A perfect square trinomial has the form ((a + b)^2 = a^2 + 2ab + b^2). Comparing the given expression (x^2 + 2xy + y^2) with the form (a^2 + 2ab + b^2), we identify (a = x) and (b = y). Substituting these values into the perfect square formula, we get:[(x + y)^2 = x^2 + 2xy + y^2]Expanding ((x + y)^2) confirms this:[(x + y)(x + y) = x^2 + xy + xy + y^2 = x^2 + 2xy + y^2]Thus, the factorized form of (x^2 + 2xy + y^2) is (boxed{(x + y)^2}).]

🔑:AnswerThe given expression can be rewritten as (x+y)^2.Therefore, the factorized form is boxed{(x+y)^2}.

🧠:[<think>Okay, let's see. Teacher Li wants to buy both type A and type B pens with 30 yuan. Each type A pen is 5 yuan, and each type B pen is 2 yuan. We need to figure out how many different purchasing plans she has. The options are A to D, with D being 1 type, which I guess means only one possible combination. Hmm.First, let's translate this into equations. Let me denote the number of type A pens as x and type B pens as y. Since she's buying both types, both x and y have to be at least 1. The total cost is 5x + 2y = 30 yuan. We need to find all possible positive integer solutions (x, y) to this equation.So, the equation is 5x + 2y = 30. Let's try to solve for y in terms of x. Subtract 5x from both sides: 2y = 30 - 5x. Then divide by 2: y = (30 - 5x)/2. Since y has to be a positive integer, (30 - 5x) must be even and positive.First, let's make sure that (30 - 5x) is even. 30 is even, and 5x is 5 times x. 5 is odd, so 5x is odd if x is odd and even if x is even. So, even minus odd is odd, and even minus even is even. Therefore, for (30 - 5x) to be even, 5x must be even, which happens when x is even. So x must be even. Therefore, x has to be an even positive integer.But also, y has to be positive. So (30 - 5x)/2 > 0. Let's solve this inequality:30 - 5x > 0 30 > 5x 6 > xSo x must be less than 6. But since x is a positive integer and even, possible values of x are 2, 4, and... wait, x can be 1, 2, 3, 4, 5. But since x has to be even, x can be 2, 4. Wait, 6 is not allowed because x must be less than 6. Wait, 6 > x, so x < 6. So x can be 1, 2, 3, 4, 5. But x must be even, so only 2, 4. Wait, but if x is even, then x can be 2, 4. Let me check:If x is 2: y = (30 - 5*2)/2 = (30 - 10)/2 = 20/2 = 10. So y = 10. That's valid.If x is 4: y = (30 - 5*4)/2 = (30 - 20)/2 = 10/2 = 5. So y = 5. Valid.What if x is 6? Then x is 6, but 6 is not less than 6, so that's invalid. So x can't be 6. How about x = 0? But the problem says she has to buy both types, so x can't be 0. Similarly, y can't be 0.Wait, but wait. What about x being 1? x is odd, so (30 - 5*1) = 25, which is odd, so y would be 25/2 = 12.5, which is not an integer. So that's invalid.Similarly, x = 3: (30 - 15)/2 = 15/2 = 7.5. Not an integer. x = 5: (30 - 25)/2 = 5/2 = 2.5. Not an integer. So the only valid x's are even numbers 2 and 4. Therefore, two possible plans: (2,10) and (4,5). So the answer should be 2 types, which is option C.Wait, but let me double-check. Maybe there's another way. Let me try x = 2: 5*2 = 10, so 30 -10 =20, which is 2y, so y=10. Correct. x=4: 5*4=20, 30-20=10, so y=5. Correct. What if x=6: 5*6=30, so y=0, which is invalid. x=0: y=15, but also invalid because she needs to buy both. So only two possibilities. Therefore, the answer is C: 2 types.But wait, let me check the problem statement again. It says "buying both types of signature pens". So x and y must both be at least 1. So, in our solutions, x=2,y=10 and x=4,y=5. So two solutions. Therefore, the answer is C:2 types. So option C is correct.But let me make sure there's no other way. For example, suppose x is even numbers less than 6. So x=2 and 4. So two plans. Therefore, answer C. Yeah.Final Answerboxed{C}</think>Teacher Li plans to buy both type A and type B pens with 30 yuan. Each type A pen costs 5 yuan, and each type B pen costs 2 yuan. We need to find the number of possible purchasing plans where both types of pens are bought.The equation for the total cost is:[ 5x + 2y = 30 ]where ( x ) and ( y ) are positive integers.Rearranging for ( y ):[ y = frac{30 - 5x}{2} ]For ( y ) to be a positive integer, ( 30 - 5x ) must be even and positive. Since 30 is even, ( 5x ) must also be even, implying ( x ) must be even. Additionally, ( 30 - 5x > 0 ) leads to:[ 30 > 5x ][ x < 6 ]Possible even values for ( x ) (since ( x ) must be a positive integer) are 2 and 4.- If ( x = 2 ): [ y = frac{30 - 5 cdot 2}{2} = frac{20}{2} = 10 ]- If ( x = 4 ): [ y = frac{30 - 5 cdot 4}{2} = frac{10}{2} = 5 ]Both solutions result in positive integers for ( y ). Therefore, there are 2 possible purchasing plans.The answer is (boxed{C}).,<think>Okay, so Teacher Li wants to buy both type A and type B pens with 30 yuan. Each type A pen costs 5 yuan, and each type B pen costs 2 yuan. The question is asking how many different purchasing plans there are, and the options are from A to D, with D being only 1 type. Let me think about how to approach this.First, let me restate the problem to make sure I understand it. Teacher Li needs to spend exactly 30 yuan on pens, buying at least one of each type. So, he can't buy all A or all B pens; he has to have a combination. The question is about how many different combinations (plans) are possible. The answer choices suggest it could be 4, 3, 2, or 1 possible plans.To solve this, I think I need to set up an equation. Let’s denote the number of type A pens as x and the number of type B pens as y. Then, the total cost would be 5x + 2y = 30 yuan. Since he has to buy both types, x and y must be at least 1.So, the equation is 5x + 2y = 30, where x ≥ 1 and y ≥ 1, and both x and y are integers. We need to find the number of integer solutions (x, y) that satisfy this equation.Let me solve for y in terms of x. Starting with 5x + 2y = 30, subtract 5x from both sides: 2y = 30 - 5x. Then, divide both sides by 2: y = (30 - 5x)/2. For y to be an integer, (30 - 5x) must be even. Since 5x is 5 times x, and 5 is odd, 5x will be odd if x is odd, and even if x is even. 30 is even, so even minus odd is odd, and even minus even is even. Therefore, for (30 - 5x) to be even, 5x must be even, so x must be even. Therefore, x has to be an even number.But also, since x ≥ 1, and y must be at least 1, we need to find the possible even values of x such that y is also a positive integer.Let me formalize this:x must be a positive integer, even, such that y = (30 - 5x)/2 ≥ 1.So, let's find the possible values of x.First, since x must be at least 1 and even, the smallest possible x is 2 (since x=1 is odd and would make 5x=5, then 30-5=25, which divided by 2 gives y=12.5, not integer). Wait, but x=1 is odd, and we concluded x must be even. Wait, but earlier reasoning: since 5x must be even, x must be even. Therefore, x can be 2, 4, 6, etc., but we need to make sure that y is also positive.So, starting with x=2:x=2:y = (30 - 5*2)/2 = (30 - 10)/2 = 20/2 = 10. So y=10. That's valid.x=4:y=(30 - 20)/2=10/2=5. y=5. Valid.x=6:y=(30 - 30)/2=0/2=0. But y must be at least 1, so y=0 is invalid. Therefore, x=6 is too large.Wait, so x can be 2, 4, but x=6 gives y=0, which is invalid. What about x=0? But x must be at least 1, so x cannot be 0.Wait, but maybe we can check for x=1:x=1:y=(30 -5)/2=25/2=12.5, which is not integer, so invalid.x=3:y=(30 -15)/2=15/2=7.5, not integer. So invalid.x=5:y=(30 -25)/2=5/2=2.5, not integer. So invalid.So only even x starting at 2, and next even is 4, then 6 which is invalid. So x can be 2 and 4. Therefore, two possible values of x, each corresponding to a unique y. So the purchasing plans are (2,10) and (4,5). Therefore, two types of purchasing plans. So the answer should be option C: 2 types.Wait, but let me double-check. Let me list all possible even x where x ≥1 and y ≥1.x must be even, so possible x values:2, 4, 6, 8, ..., but with y ≥1.So, let's write the formula:y = (30 -5x)/2 ≥1So, (30 -5x)/2 ≥1Multiply both sides by 2: 30 -5x ≥2Subtract 30: -5x ≥ -28Multiply both sides by -1 (and reverse inequality): 5x ≤28Therefore, x ≤28/5 =5.6Since x must be even and positive integer, x can be 2,4 (since 6 would be 6, which is greater than 5.6). Therefore, only x=2 and x=4. Hence, two solutions. Therefore, the answer is 2 types, which is option C.Wait, but let me check once more. Maybe I made a mistake in considering only even x.But according to the equation y=(30−5x)/2, for y to be integer, 30−5x must be even. Since 30 is even and 5x is even or odd based on x. 5x is even if x is even. Therefore, 30−even is even. If x is odd, 5x is odd, so 30−odd is odd, which would make y non-integer. Therefore, x must be even. So x must be even. Therefore, possible x's are 2,4,6,... but as above x<=5.6, so x=2,4. So two solutions.Hence, answer is C:2 types.Wait, but let's check if there's another way. Suppose we let x be any integer >=1, and then check if y is integer and >=1.For x=1: y=(30−5)/2=25/2=12.5 Not integer.x=2: y=10. Integer. Valid.x=3: 15/2=7.5 Not integer.x=4: 5. Integer. Valid.x=5: (30-25)/2=2.5 Not integer.x=6: 0.0. But y=0, which is invalid.Therefore, only x=2 and x=4. So two purchasing plans. So answer C.Alternatively, maybe the problem is that when x=6, y=0, but since he must buy both types, y=0 is invalid, so x can't be 6. Therefore, indeed, only two options. Therefore, answer is C.Final Answerboxed{C}</think>Teacher Li plans to buy both types of signature pens A and B with 30 yuan. Each type A pen costs 5 yuan, and each type B pen costs 2 yuan. We need to find the number of different purchasing plans where both types of pens are bought.We set up the equation for the total cost:[ 5x + 2y = 30 ]where ( x ) and ( y ) are the number of type A and type B pens, respectively, and both ( x ) and ( y ) must be at least 1.Solving for ( y ) in terms of ( x ):[ y = frac{30 - 5x}{2} ]For ( y ) to be an integer, ( 30 - 5x ) must be even. Since 30 is even, ( 5x ) must also be even, implying ( x ) must be even. Additionally, ( y ) must be at least 1.Checking possible even values of ( x ):- When ( x = 2 ): [ y = frac{30 - 5 times 2}{2} = frac{20}{2} = 10 ] This is valid as ( y = 10 ).- When ( x = 4 ): [ y = frac{30 - 5 times 4}{2} = frac{10}{2} = 5 ] This is valid as ( y = 5 ).- When ( x = 6 ): [ y = frac{30 - 5 times 6}{2} = frac{0}{2} = 0 ] This is invalid as ( y = 0 ).Thus, the valid values of ( x ) are 2 and 4, corresponding to ( y = 10 ) and ( y = 5 ) respectively. Therefore, there are 2 types of purchasing plans.[boxed{C}]]

🔑:To solve this problem, let's denote the number of type A pens Teacher Li buys as x and the number of type B pens as y. Given that each type A pen costs 5 yuan and each type B pen costs 2 yuan, and Teacher Li plans to spend exactly 30 yuan, we can set up the following equation to represent this situation:[5x + 2y = 30]To find the possible purchasing plans, we need to solve for x and y under the constraint that both x and y must be positive integers because Teacher Li cannot buy a fraction of a pen. Rearranging the equation for x, we get:[x = frac{30 - 2y}{5}]This equation tells us that for any given number of type B pens (y), we can calculate the number of type A pens (x) that Teacher Li can buy. However, since x must be a positive integer, the right-hand side of the equation must also be an integer. This means that 30 - 2y must be divisible by 5.By testing values of y that are positive integers, we find that there are specific combinations of x and y that satisfy the equation:1. When y = 5, we substitute into the equation to get x = frac{30 - 2(5)}{5} = frac{30 - 10}{5} = frac{20}{5} = 4. So, one plan is to buy 4 type A pens and 5 type B pens.2. When y = 10, substituting into the equation gives x = frac{30 - 2(10)}{5} = frac{30 - 20}{5} = frac{10}{5} = 2. Thus, another plan is to buy 2 type A pens and 10 type B pens.These are the only combinations of x and y that satisfy the given conditions and result in positive integers for both variables. Therefore, Teacher Li has exactly 2 different purchasing plans available.Hence, the correct answer is boxed{C}.