| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we can use Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots. For a quadratic equation of the form \(ax^2 + bx + c = 0\), if \(m\) and \(n\) are its roots, then: 1. \(m + n = -\frac{b}{a}\) 2. \(mn = \frac{c}{a}\) Given the quadratic equation \(2x^2 + 15x + 16 = 0\), we have \(a = 2\), \(b = 15\), and \(c = 16\). According to Vieta's formulas: 1. \(m + n = -\frac{15}{2}\) 2. \(mn = \frac{16}{2} = 8\) We are asked to find the value of \(\frac{1}{m} + \frac{1}{n}\). This can be rewritten as: \[ \frac{1}{m} + \frac{1}{n} = \frac{m + n}{mn} \] Substituting the values from Vieta's formulas: \[ \frac{1}{m} + \frac{1}{n} = \frac{-\frac{15}{2}}{8} = -\frac{15}{16} \] Therefore, the value of \(\frac{1}{m} + \frac{1}{n}\) is \(-\frac{15}{16}\). |
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| deepseekr1 | Given the quadratic equation \(2x^2 + 15x + 16 = 0\) with roots \(m\) and \(n\), we need to find the value of \(\frac{1}{m} + \frac{1}{n}\). Using Vieta's formulas, we know that for a quadratic equation \(ax^2 + bx + c = 0\), the sum of the roots \(m + n\) is given by \(-\frac{b}{a}\) and the product of the roots \(mn\) is given by \(\frac{c}{a}\). For the given equation \(2x^2 + 15x + 16 = 0\): - The sum of the roots \(m + n = -\frac{15}{2}\) - The product of the roots \(mn = \frac{16}{2} = 8\) We can express \(\frac{1}{m} + \frac{1}{n}\) as \(\frac{m + n}{mn}\) by combining the fractions. Substituting the values from Vieta's formulas: \[ \frac{1}{m} + \frac{1}{n} = \frac{m + n}{mn} = \frac{-\frac{15}{2}}{8} = -\frac{15}{2} \times \frac{1}{8} = -\frac{15}{16} \] Thus, the value of \(\frac{1}{m} + \frac{1}{n}\) is \(\boxed{-\dfrac{15}{16}}\). |
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| gpto1 | To find the value of \(\dfrac{1}{m} + \dfrac{1}{n}\), where \(m\) and \(n\) are the roots of the quadratic equation \(2x^2 + 15x + 16 = 0\), we'll use Vieta's formulas, which relate the sums and products of the roots to the coefficients of the quadratic equation. **Step 1: Identify the coefficients of the quadratic equation.** The general form of a quadratic equation is: \[ ax^2 + bx + c = 0 \] For the given equation: \[ 2x^2 + 15x + 16 = 0 \] - \(a = 2\) - \(b = 15\) - \(c = 16\) **Step 2: Use Vieta's formulas to find the sum and product of the roots.** Vieta's formulas state that for a quadratic equation \(ax^2 + bx + c = 0\) with roots \(m\) and \(n\): 1. The sum of the roots: \[ m + n = -\dfrac{b}{a} \] 2. The product of the roots: \[ mn = \dfrac{c}{a} \] Calculate the sum and product: 1. Sum of the roots: \[ m + n = -\dfrac{b}{a} = -\dfrac{15}{2} \] 2. Product of the roots: \[ mn = \dfrac{c}{a} = \dfrac{16}{2} = 8 \] **Step 3: Find \(\dfrac{1}{m} + \dfrac{1}{n}\) using the sum and product.** The expression \(\dfrac{1}{m} + \dfrac{1}{n}\) can be combined over a common denominator: \[ \dfrac{1}{m} + \dfrac{1}{n} = \dfrac{n + m}{mn} \] Substitute the values of \(m + n\) and \(mn\): \[ \dfrac{1}{m} + \dfrac{1}{n} = \dfrac{-(15/2)}{8} \] Simplify the numerator and denominator: 1. Multiply the numerator and denominator by 2 to eliminate the fraction in the numerator: \[ \dfrac{-(15/2)}{8} = \dfrac{-15}{2 \times 8} = \dfrac{-15}{16} \] **Answer:** \[ \dfrac{1}{m} + \dfrac{1}{n} = -\dfrac{15}{16} \] |
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| GPT-4o | To find the value of \( \frac{1}{m} + \frac{1}{n} \) where \( m \) and \( n \) are the roots of the quadratic equation \( 2x^2 + 15x + 16 = 0 \), we can use Vieta's formulas, which relate the coefficients of the polynomial to sums and products of the roots. For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the sum of the roots \( m \) and \( n \) is given by: \[ m + n = -\frac{b}{a} \] The product of the roots \( m \) and \( n \) is given by: \[ mn = \frac{c}{a} \] In the given equation \( 2x^2 + 15x + 16 = 0 \): - \( a = 2 \) - \( b = 15 \) - \( c = 16 \) Applying Vieta's formulas, we get: \[ m + n = -\frac{15}{2} \] \[ mn = \frac{16}{2} = 8 \] Now, we need to find \( \frac{1}{m} + \frac{1}{n} \). We can use the identity: \[ \frac{1}{m} + \frac{1}{n} = \frac{m + n}{mn} \] Substituting the values we found using Vieta's formulas: \[ \frac{1}{m} + \frac{1}{n} = \frac{m + n}{mn} = \frac{-\frac{15}{2}}{8} = \frac{-15}{2 \cdot 8} = \frac{-15}{16} \] Therefore, the value of \( \frac{1}{m} + \frac{1}{n} \) is: \[ \frac{1}{m} + \frac{1}{n} = -\frac{15}{16} \] |
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| llama3.1-70B-Instruct | To find the value of 1/m + 1/n, we can use the fact that m and n are the roots of the equation 2x^2 + 15x + 16 = 0. First, we can write the equation in the form: 2x^2 + 15x + 16 = 0 We can then factor the left-hand side of the equation: (2x + 8)(x + 2) = 0 This tells us that either (2x + 8) = 0 or (x + 2) = 0. Solving for x, we get: 2x + 8 = 0 --> 2x = -8 --> x = -4 x + 2 = 0 --> x = -2 So, the roots of the equation are m = -4 and n = -2. Now, we can find the value of 1/m + 1/n: 1/m + 1/n = 1/(-4) + 1/(-2) = -1/4 - 1/2 = -3/4 Therefore, the value of 1/m + 1/n is -3/4. |
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| llama3.1-405B-Instruct | To find the value of 1/m + 1/n, we can use the fact that m and n are the roots of the equation 2x^2 + 15x + 16 = 0. First, we can factor the quadratic expression: 2x^2 + 15x + 16 = (2x + 8)(x + 2) = 0 This tells us that either (2x + 8) = 0 or (x + 2) = 0. Solving for x, we get: 2x + 8 = 0 --> 2x = -8 --> x = -4 x + 2 = 0 --> x = -2 So, the roots of the equation are m = -4 and n = -2. Now, we can find the value of 1/m + 1/n: 1/m + 1/n = 1/(-4) + 1/(-2) = -1/4 - 1/2 = -3/4 Therefore, the value of 1/m + 1/n is -3/4. |
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| DeepSeek-Coder-V2 | To solve the problem, we need to find the value of 1/m + 1/n, where m and n are the roots of the quadratic equation 2x^2 + 15x + 16 = 0. First, let's use the quadratic formula to find the roots of the equation. The quadratic formula is given by: x = (-b ± sqrt(b^2 - 4ac)) / (2a) For the given equation 2x^2 + 15x + 16 = 0, the coefficients are: a = 2, b = 15, and c = 16. Now, let's plug these values into the quadratic formula: x = (-15 ± sqrt(15^2 - 4*2*16)) / (2*2) x = (-15 ± sqrt(225 - 128)) / 4 x = (-15 ± sqrt(97)) / 4 So, the roots of the equation are: m = (-15 + sqrt(97)) / 4 n = (-15 - sqrt(97)) / 4 Now, we need to find the value of 1/m + 1/n. We can use the relationship between the roots and the coefficients of the quadratic equation. According to Vieta's formulas, for a quadratic equation ax^2 + bx + c = 0, the sum of the roots (m + n) is -b/a, and the product of the roots (mn) is c/a. For our equation 2x^2 + 15x + 16 = 0: m + n = -b/a = -15/2 mn = c/a = 16/2 = 8 Now, let's find 1/m + 1/n: 1/m + 1/n = (m + n) / mn Substitute the values we found using Vieta's formulas: 1/m + 1/n = (-15/2) / 8 1/m + 1/n = -15 / (2 * 8) 1/m + 1/n = -15 / 16 So, the value of 1/m + 1/n is -15/16. |
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| Qwen2-72B-Instruct | To solve this problem, we can use Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots. For a quadratic equation of the form \(ax^2 + bx + c = 0\), if \(m\) and \(n\) are its roots, then: 1. \(m + n = -\frac{b}{a}\) 2. \(mn = \frac{c}{a}\) Given the quadratic equation \(2x^2 + 15x + 16 = 0\), we have \(a = 2\), \(b = 15\), and \(c = 16\). According to Vieta's formulas: 1. \(m + n = -\frac{15}{2}\) 2. \(mn = \frac{16}{2} = 8\) We are asked to find the value of \(\frac{1}{m} + \frac{1}{n}\). This can be rewritten as: \[ \frac{1}{m} + \frac{1}{n} = \frac{m + n}{mn} \] Substituting the values from Vieta's formulas: \[ \frac{1}{m} + \frac{1}{n} = \frac{-\frac{15}{2}}{8} = -\frac{15}{16} \] Therefore, the value of \(\frac{1}{m} + \frac{1}{n}\) is \(-\frac{15}{16}\). |
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| baichuan4 | To find the value of 1/m + 1/n, where m and n are the roots of the equation 2x^2 + 15x + 16 = 0, we can use the properties of quadratic equations. First, let's write the quadratic equation in standard form: 2x^2 + 15x + 16 = 0 We know that for a quadratic equation ax^2 + bx + c = 0, the sum of the roots (m + n) is given by -b/a, and the product of the roots (mn) is given by c/a. For our equation, a = 2, b = 15, and c = 16. So, m + n = -b/a = -15/2 And mn = c/a = 16/2 = 8 Now, we want to find the value of 1/m + 1/n. We can write this as: 1/m + 1/n = (n + m) / (mn) Substituting the values we found earlier: 1/m + 1/n = (-15/2) / 8 = -15 / 16 Therefore, the value of 1/m + 1/n is -15/16. |
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| doubao | For a quadratic equation ax^2 + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a. In the given equation 2x^2 + 15x + 16 = 0, a = 2, b = 15, c = 16. The sum of the roots m and n is -15/2 and the product is 16/2 = 8. 1/m + 1/n = (n + m)/(mn) = (-15/2) / 8 = -15/16. |
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| minimax | To find the value of 1/m + 1/n, where m and n are the roots of the quadratic equation 2x^2 + 15x + 16 = 0, we can use Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots. For a quadratic equation of the form ax^2 + bx + c = 0, where a, b, and c are constants, and m and n are the roots, Vieta's formulas tell us that: m + n = -b/a mn = c/a In our case, the equation is 2x^2 + 15x + 16 = 0, so a = 2, b = 15, and c = 16. Using Vieta's formulas, we can find the sum and product of the roots: m + n = -b/a = -15/2 mn = c/a = 16/2 = 8 Now, we want to find the value of 1/m + 1/n. We can rewrite this expression in terms of the sum and product of the roots: 1/m + 1/n = (n + m) / (mn) We already know m + n and mn, so we can substitute these values into the expression: 1/m + 1/n = (-15/2) / 8 1/m + 1/n = -15 / (2 * 8) 1/m + 1/n = -15 / 16 Therefore, the value of 1/m + 1/n is -15/16. |
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| kimi | To find the value of \( \frac{1}{m} + \frac{1}{n} \) where \( m \) and \( n \) are the roots of the quadratic equation \( 2x^2 + 15x + 16 = 0 \), we can use Vieta's formulas. Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the sum of the roots \( m + n \) is given by \( -\frac{b}{a} \) and the product of the roots \( mn \) is given by \( \frac{c}{a} \). In our case, the equation is \( 2x^2 + 15x + 16 = 0 \), so: - \( a = 2 \) - \( b = 15 \) - \( c = 16 \) Using Vieta's formulas, we get: \[ m + n = -\frac{b}{a} = -\frac{15}{2} \] \[ mn = \frac{c}{a} = \frac{16}{2} = 8 \] Now, we want to find \( \frac{1}{m} + \frac{1}{n} \). This can be rewritten as: \[ \frac{1}{m} + \frac{1}{n} = \frac{n + m}{mn} \] Substituting the values we found from Vieta's formulas: \[ \frac{1}{m} + \frac{1}{n} = \frac{-\frac{15}{2}}{8} \] Now, we simplify the fraction: \[ \frac{1}{m} + \frac{1}{n} = -\frac{15}{16} \] So, the value of \( \frac{1}{m} + \frac{1}{n} \) is \( -\frac{15}{16} \). |
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| wenxinyiyan | 首先，我们知道一元二次方程 $ax^2 + bx + c = 0$ 的两个根 $m$ 和 $n$ 满足以下关系： $m + n = -\frac{b}{a}$ $mn = \frac{c}{a}$ 对于给定的方程 $2x^2 + 15x + 16 = 0$，我们有 $a = 2, b = 15, c = 16$。 根据上面的关系，我们可以得到： $m + n = -\frac{15}{2}$ $mn = \frac{16}{2} = 8$ 接下来，我们需要求 $\frac{1}{m} + \frac{1}{n}$。 根据分数的加法公式，我们有： $\frac{1}{m} + \frac{1}{n} = \frac{m + n}{mn}$ 将 $m + n = -\frac{15}{2}$ 和 $mn = 8$ 代入上式，得到： $\frac{1}{m} + \frac{1}{n} = \frac{-\frac{15}{2}}{8} = -\frac{15}{16}$ 故答案为：$-\frac{15}{16}$。 |
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| zhipu | To find the value of \( \frac{1}{m} + \frac{1}{n} \), where \( m \) and \( n \) are the roots of the quadratic equation \( 2x^2 + 15x + 16 = 0 \), we can use the properties of the roots of quadratic equations. For a quadratic equation of the form \( ax^2 + bx + c = 0 \), if \( m \) and \( n \) are the roots, then according to Vieta's formulas: 1. The sum of the roots \( m + n = -\frac{b}{a} \) 2. The product of the roots \( mn = \frac{c}{a} \) In our equation \( 2x^2 + 15x + 16 = 0 \), we have \( a = 2 \), \( b = 15 \), and \( c = 16 \). Applying Vieta's formulas, we get: 1. \( m + n = -\frac{15}{2} \) 2. \( mn = \frac{16}{2} = 8 \) Now, we want to find \( \frac{1}{m} + \frac{1}{n} \). This can be expressed as: \[ \frac{1}{m} + \frac{1}{n} = \frac{m + n}{mn} \] Substituting the values from Vieta's formulas: \[ \frac{1}{m} + \frac{1}{n} = \frac{-\frac{15}{2}}{8} \] To simplify this, we multiply the numerator and the denominator by 2 to get rid of the fraction in the numerator: \[ \frac{1}{m} + \frac{1}{n} = \frac{-15}{2 \cdot 8} = \frac{-15}{16} \] Therefore, the value of \( \frac{1}{m} + \frac{1}{n} \) is \( -\frac{15}{16} \). |
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| gpto1 | To find the value of \(\dfrac{1}{m} + \dfrac{1}{n}\), where \(m\) and \(n\) are the roots of the quadratic equation \(2x^2 + 15x + 16 = 0\), we'll use Vieta's formulas, which relate the sums and products of the roots to the coefficients of the quadratic equation. **Step 1: Identify the coefficients of the quadratic equation.** The general form of a quadratic equation is: \[ ax^2 + bx + c = 0 \] For the given equation: \[ 2x^2 + 15x + 16 = 0 \] - \(a = 2\) - \(b = 15\) - \(c = 16\) **Step 2: Use Vieta's formulas to find the sum and product of the roots.** Vieta's formulas state that for a quadratic equation \(ax^2 + bx + c = 0\) with roots \(m\) and \(n\): 1. The sum of the roots: \[ m + n = -\dfrac{b}{a} \] 2. The product of the roots: \[ mn = \dfrac{c}{a} \] Calculate the sum and product: 1. Sum of the roots: \[ m + n = -\dfrac{b}{a} = -\dfrac{15}{2} \] 2. Product of the roots: \[ mn = \dfrac{c}{a} = \dfrac{16}{2} = 8 \] **Step 3: Find \(\dfrac{1}{m} + \dfrac{1}{n}\) using the sum and product.** The expression \(\dfrac{1}{m} + \dfrac{1}{n}\) can be combined over a common denominator: \[ \dfrac{1}{m} + \dfrac{1}{n} = \dfrac{n + m}{mn} \] Substitute the values of \(m + n\) and \(mn\): \[ \dfrac{1}{m} + \dfrac{1}{n} = \dfrac{-(15/2)}{8} \] Simplify the numerator and denominator: 1. Multiply the numerator and denominator by 2 to eliminate the fraction in the numerator: \[ \dfrac{-(15/2)}{8} = \dfrac{-15}{2 \times 8} = \dfrac{-15}{16} \] **Answer:** \[ \dfrac{1}{m} + \dfrac{1}{n} = -\dfrac{15}{16} \] |
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| deepseekv3 | To find the value of \( \frac{1}{m} + \frac{1}{n} \), where \( m \) and \( n \) are the roots of the quadratic equation \( 2x^2 + 15x + 16 = 0 \), we can follow these steps: ### Step 1: Identify the Coefficients The given quadratic equation is: \[ 2x^2 + 15x + 16 = 0 \] For a general quadratic equation \( ax^2 + bx + c = 0 \): - Sum of the roots (\( m + n \)) = \( -\frac{b}{a} \) - Product of the roots (\( m \times n \)) = \( \frac{c}{a} \) From the given equation: - \( a = 2 \) - \( b = 15 \) - \( c = 16 \) ### Step 2: Calculate the Sum and Product of the Roots \[ m + n = -\frac{b}{a} = -\frac{15}{2} \] \[ m \times n = \frac{c}{a} = \frac{16}{2} = 8 \] ### Step 3: Compute \( \frac{1}{m} + \frac{1}{n} \) \[ \frac{1}{m} + \frac{1}{n} = \frac{n + m}{m \times n} \] Substitute the values from Step 2: \[ \frac{1}{m} + \frac{1}{n} = \frac{-\frac{15}{2}}{8} = -\frac{15}{16} \] ### Final Answer \[ \boxed{-\dfrac{15}{16}} \] |
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