| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To complete the square for the given quadratic expression \(\frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}}\), we will follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms:** \[ \frac{34}{\sqrt{5}} x^2 - 8 \sqrt{5} x + \frac{7}{\sqrt{5}} = \frac{34}{\sqrt{5}} \left( x^2 - \frac{8 \sqrt{5}}{34} x \right) + \frac{7}{\sqrt{5}} \] 2. **Simplify the coefficient inside the parentheses:** \[ = \frac{34}{\sqrt{5}} \left( x^2 - \frac{40}{34} x \right) + \frac{7}{\sqrt{5}} \] \[ = \frac{34}{\sqrt{5}} \left( x^2 - \frac{20}{17} x \right) + \frac{7}{\sqrt{5}} \] 3. **Complete the square inside the parentheses:** To complete the square, we need to add and subtract the square of half the coefficient of \(x\). The coefficient of \(x\) is \(-\frac{20}{17}\), so half of this is \(-\frac{10}{17}\). Squaring this gives \(\left( -\frac{10}{17} \right)^2 = \frac{100}{289}\). \[ = \frac{34}{\sqrt{5}} \left( x^2 - \frac{20}{17} x + \frac{100}{289} - \frac{100}{289} \right) + \frac{7}{\sqrt{5}} \] \[ = \frac{34}{\sqrt{5}} \left[ \left( x - \frac{10}{17} \right)^2 - \frac{100}{289} \right] + \frac{7}{\sqrt{5}} \] 4. **Distribute the \(\frac{34}{\sqrt{5}}\) and simplify:** \[ = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{34}{\sqrt{5}} \cdot \frac{100}{289} + \frac{7}{\sqrt{5}} \] \[ = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{3400}{289 \sqrt{5}} + \frac{7}{\sqrt{5}} \] 5. **Combine the constant terms:** To combine the constants, we find a common denominator: \[ = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 + \left( \frac{7 \cdot 289 - 3400}{289 \sqrt{5}} \right) \] \[ = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 + \left( \frac{2023 - 3400}{289 \sqrt{5}} \right) \] \[ = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 + \left( \frac{-1377}{289 \sqrt{5}} \right) \] Thus, the completed square form of the given quadratic is: \[ \boxed{\frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{1377}{289 \sqrt{5}}} \] |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| deepseek-r1 | To complete the square for the quadratic expression \(\frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}}\), we start by factoring out \(\frac{1}{\sqrt{5}}\) from each term: \[ \frac{1}{\sqrt{5}} \left( 34x^2 - 40x + 7 \right) \] Next, we complete the square for the quadratic expression inside the brackets. We factor out the coefficient of \(x^2\) from the first two terms: \[ 34 \left( x^2 - \frac{20}{17}x \right) + 7 \] To complete the square, we take half of the coefficient of \(x\), which is \(\frac{10}{17}\), and square it to get \(\frac{100}{289}\). We add and subtract this value inside the parentheses: \[ 34 \left( \left( x - \frac{10}{17} \right)^2 - \frac{100}{289} \right) + 7 \] Distributing the 34 and simplifying the constant terms: \[ 34 \left( x - \frac{10}{17} \right)^2 - \frac{3400}{289} + 7 \] Converting 7 to a fraction with denominator 289: \[ 34 \left( x - \frac{10}{17} \right)^2 - \frac{3400}{289} + \frac{2023}{289} \] Combining the constants: \[ 34 \left( x - \frac{10}{17} \right)^2 - \frac{1377}{289} \] Simplifying \(\frac{1377}{289}\) to \(\frac{81}{17}\): \[ 34 \left( x - \frac{10}{17} \right)^2 - \frac{81}{17} \] Now, we multiply by \(\frac{1}{\sqrt{5}}\) and rationalize the denominators: \[ \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{81}{17 \sqrt{5}} \] Rationalizing each term: \[ \frac{34 \sqrt{5}}{5} \left( x - \frac{10}{17} \right)^2 - \frac{81 \sqrt{5}}{85} \] Thus, the final answer is: \[ \boxed{\dfrac{34\sqrt{5}}{5}\left(x - \dfrac{10}{17}\right)^2 - \dfrac{81\sqrt{5}}{85}} \] |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| gpto1 | To complete the square for the quadratic expression: \[ Q(x) = \frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}} \] We can rationalize the denominators and bring all terms under a common denominator: 1. **Rationalize Denominators:** \[ \frac{34 x^2}{\sqrt{5}} = \frac{34 x^2 \sqrt{5}}{5} \] \[ \frac{7}{\sqrt{5}} = \frac{7 \sqrt{5}}{5} \] 2. **Rewrite the Expression:** \[ Q(x) = \frac{34 x^2 \sqrt{5}}{5} - 8 \sqrt{5} x + \frac{7 \sqrt{5}}{5} \] Factor out \(\sqrt{5}\) from all terms: \[ Q(x) = \sqrt{5} \left( \frac{34 x^2}{5} - 8 x + \frac{7}{5} \right) \] 3. **Complete the Square Inside the Parentheses:** a. **Multiply Through by 5 to Eliminate Fractions:** \[ 5 \left( \frac{34 x^2}{5} - 8 x + \frac{7}{5} \right) = 34 x^2 - 40 x + 7 \] b. **Factor Out Coefficient of \( x^2 \):** \[ 34 x^2 - 40 x + 7 = 34 \left( x^2 - \frac{20}{17} x \right) + 7 \] c. **Complete the Square:** - Compute \(\left( \frac{20}{34} \right)^2 = \left( \frac{10}{17} \right)^2 = \frac{100}{289}\) - Add and subtract this inside the parentheses: \[ 34 \left[ \left( x - \frac{10}{17} \right)^2 - \frac{100}{289} \right] + 7 \] - Simplify constants: \[ 34 \left( x - \frac{10}{17} \right)^2 - \frac{200}{17} + 7 \] d. **Combine Constants:** \[ - \frac{200}{17} + 7 = - \frac{200}{17} + \frac{119}{17} = -\frac{81}{17} \] e. **Final Completed Square Form:** \[ Q(x) = \sqrt{5} \left[ \frac{34}{5} \left( x - \frac{10}{17} \right)^2 - \frac{81}{85} \right] \] 4. **Express the Final Answer:** \[ Q(x) = \frac{34 \sqrt{5}}{5} \left( x - \frac{10}{17} \right)^2 - \frac{81 \sqrt{5}}{85} \] **Final Answer:** \[ \boxed{\displaystyle \frac{34\sqrt{5}}{5}\left(x-\dfrac{10}{17}\right)^2\ -\ \dfrac{81\sqrt{5}}{85}} \] |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| gpt-4o-2024-08-06 | To complete the square for the quadratic expression \(\frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}}\), we start by focusing on the quadratic and linear terms: \(\frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x\). First, factor out the coefficient of \(x^2\) from these two terms: \[ \frac{34}{\sqrt{5}} \left( x^2 - \frac{8 \sqrt{5}}{\frac{34}{\sqrt{5}}} x \right) \] Simplify the coefficient of \(x\): \[ \frac{8 \sqrt{5}}{\frac{34}{\sqrt{5}}} = \frac{8 \sqrt{5} \cdot \sqrt{5}}{34} = \frac{40}{34} = \frac{20}{17} \] So, the expression becomes: \[ \frac{34}{\sqrt{5}} \left( x^2 - \frac{20}{17} x \right) \] Next, complete the square inside the parentheses. Take half of the coefficient of \(x\), square it, and add and subtract it inside the parentheses: \[ \left( \frac{20}{34} \right)^2 = \left( \frac{10}{17} \right)^2 = \frac{100}{289} \] Add and subtract \(\frac{100}{289}\) inside the parentheses: \[ x^2 - \frac{20}{17} x = \left( x - \frac{10}{17} \right)^2 - \frac{100}{289} \] Substitute back into the expression: \[ \frac{34}{\sqrt{5}} \left( \left( x - \frac{10}{17} \right)^2 - \frac{100}{289} \right) \] Distribute \(\frac{34}{\sqrt{5}}\): \[ \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{34}{\sqrt{5}} \cdot \frac{100}{289} \] Simplify the second term: \[ \frac{34 \cdot 100}{\sqrt{5} \cdot 289} = \frac{3400}{289 \sqrt{5}} \] Now, add the constant term \(\frac{7}{\sqrt{5}}\) back: \[ \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{3400}{289 \sqrt{5}} + \frac{7}{\sqrt{5}} \] Combine the constant terms: \[ \frac{7}{\sqrt{5}} - \frac{3400}{289 \sqrt{5}} = \frac{7 \cdot 289 - 3400}{289 \sqrt{5}} = \frac{2023 - 3400}{289 \sqrt{5}} = \frac{-1377}{289 \sqrt{5}} \] Thus, the completed square form of the quadratic is: \[ \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 + \frac{-1377}{289 \sqrt{5}} \] The final answer is: \[ \boxed{\frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 + \frac{-1377}{289 \sqrt{5}}} \] |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| DeepSeek-Coder-V2-Lite-Instruct | To complete the square for the quadratic expression \(\frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}}\), we will follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms:** \[ \frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x = \frac{34}{\sqrt{5}} \left( x^2 - \frac{8 \sqrt{5}}{34} x \right) = \frac{34}{\sqrt{5}} \left( x^2 - \frac{40}{34} x \right) = \frac{34}{\sqrt{5}} \left( x^2 - \frac{20}{17} x \right) \] 2. **Complete the square inside the parentheses:** \[ x^2 - \frac{20}{17} x \] To complete the square, we add and subtract \(\left( \frac{20}{34} \right)^2 = \left( \frac{10}{17} \right)^2\): \[ x^2 - \frac{20}{17} x + \left( \frac{10}{17} \right)^2 - \left( \frac{10}{17} \right)^2 = \left( x - \frac{10}{17} \right)^2 - \left( \frac{10}{17} \right)^2 \] 3. **Substitute back into the expression:** \[ \frac{34}{\sqrt{5}} \left( \left( x - \frac{10}{17} \right)^2 - \left( \frac{10}{17} \right)^2 \right) \] 4. **Simplify the expression:** \[ \frac{34}{\sqrt{5}} \left( \left( x - \frac{10}{17} \right)^2 - \frac{100}{289} \right) = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{34}{\sqrt{5}} \cdot \frac{100}{289} \] \[ = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{3400}{289 \sqrt{5}} \] 5. **Add the constant term \(\frac{7}{\sqrt{5}}\):** \[ \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{3400}{289 \sqrt{5}} + \frac{7}{\sqrt{5}} \] Combine the constants: \[ \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 + \left( \frac{7}{\sqrt{5}} - \frac{3400}{289 \sqrt{5}} \right) \] \[ = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 + \frac{7 \cdot 289 - 3400}{289 \sqrt{5}} \] \[ = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 + \frac{2023 - 3400}{289 \sqrt{5}} \] \[ = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{1377}{289 \sqrt{5}} \] 6. **Simplify the final expression:** \[ \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{1377}{289 \sqrt{5}} \] Thus, the completed square form of the quadratic expression is: \[ \boxed{\frac{34 \left( x - \frac{10}{17} \right)^2}{\sqrt{5}} - \frac{1377}{289 \sqrt{5}}} \] |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the quadratic equation The given quadratic equation is $\frac{34 x^2}{\sqrt{5}}-8 \sqrt{5} x+\frac{7}{\sqrt{5}}$. ## Step 2: Rewrite the quadratic equation in standard form To complete the square, we need to rewrite the quadratic equation in the standard form $ax^2 + bx + c$. We can achieve this by multiplying the entire equation by $\sqrt{5}$ to get rid of the fractions: $34x^2 - 40x + 7$. ## Step 3: Move the constant term to the right-hand side Move the constant term $7$ to the right-hand side of the equation: $34x^2 - 40x = -7$. ## Step 4: Divide the entire equation by the coefficient of $x^2$ Divide the entire equation by $34$ to make the coefficient of $x^2$ equal to $1$: $x^2 - \frac{40}{34}x = -\frac{7}{34}$. ## Step 5: Find the value to complete the square To complete the square, we need to add $(\frac{b}{2})^2$ to both sides of the equation, where $b$ is the coefficient of $x$. In this case, $b = -\frac{40}{34}$. So, we need to add $(\frac{-40/34}{2})^2 = (\frac{-20}{34})^2 = \frac{400}{1156}$ to both sides. ## Step 6: Add the value to both sides of the equation Add $\frac{400}{1156}$ to both sides of the equation: $x^2 - \frac{40}{34}x + \frac{400}{1156} = -\frac{7}{34} + \frac{400}{1156}$. ## Step 7: Simplify the right-hand side Simplify the right-hand side of the equation: $-\frac{7}{34} + \frac{400}{1156} = -\frac{7}{34} + \frac{100}{289} = -\frac{203}{289} + \frac{100}{289} = -\frac{103}{289}$. ## Step 8: Factor the left-hand side Factor the left-hand side of the equation: $(x - \frac{20}{34})^2 = -\frac{103}{289}$. ## Step 9: Simplify the left-hand side Simplify the left-hand side of the equation: $(x - \frac{10}{17})^2 = -\frac{103}{289}$. ## Step 10: Write the final answer Since we are asked to put our final answer in a $\boxed{}$, we will rewrite the equation as $\frac{34 x^2}{\sqrt{5}}-8 \sqrt{5} x+\frac{7}{\sqrt{5}} = \frac{34}{\sqrt{5}}(x - \frac{10}{17})^2 -\frac{103}{\sqrt{5}}$ and then put it in the box. The final answer is: $\boxed{\frac{34}{\sqrt{5}}(x - \frac{10}{17})^2 -\frac{103}{\sqrt{5}}}$. |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the coefficients of the quadratic equation The given quadratic equation is $\frac{34 x^2}{\sqrt{5}}-8 \sqrt{5} x+\frac{7}{\sqrt{5}}$. The coefficients are $a = \frac{34}{\sqrt{5}}$, $b = -8\sqrt{5}$, and $c = \frac{7}{\sqrt{5}}$. ## Step 2: Calculate the value to complete the square To complete the square, we need to calculate the value of $\left(\frac{b}{2}\right)^2$. Substituting the value of $b$, we get $\left(\frac{-8\sqrt{5}}{2}\right)^2 = (-4\sqrt{5})^2 = 16 \cdot 5 = 80$. ## Step 3: Rewrite the quadratic equation in completed square form The completed square form of the quadratic equation is $a(x-h)^2 + k$, where $h = -\frac{b}{2a}$ and $k = c - \frac{b^2}{4a}$. We already have the value of $\frac{b^2}{4a}$ from step 2, which is $80$. Now, we need to calculate $h$ and $k$. ## Step 4: Calculate the value of h $h = -\frac{b}{2a} = -\frac{-8\sqrt{5}}{2 \cdot \frac{34}{\sqrt{5}}} = \frac{8\sqrt{5}}{\frac{68}{\sqrt{5}}} = \frac{8\sqrt{5} \cdot \sqrt{5}}{68} = \frac{40}{68} = \frac{10}{17}$. ## Step 5: Calculate the value of k $k = c - \frac{b^2}{4a} = \frac{7}{\sqrt{5}} - 80 = \frac{7}{\sqrt{5}} - \frac{80\sqrt{5}}{\sqrt{5}} = \frac{7 - 80\sqrt{5}}{\sqrt{5}}$. ## Step 6: Write the completed square form of the quadratic equation The completed square form of the quadratic equation is $\frac{34}{\sqrt{5}}\left(x-\frac{10}{17}\right)^2 + \frac{7 - 80\sqrt{5}}{\sqrt{5}}$. ## Step 7: Simplify the expression $\frac{34}{\sqrt{5}}\left(x-\frac{10}{17}\right)^2 + \frac{7 - 80\sqrt{5}}{\sqrt{5}} = \frac{34}{\sqrt{5}}\left(x-\frac{10}{17}\right)^2 + \frac{7}{\sqrt{5}} - \frac{80\sqrt{5}}{\sqrt{5}} = \frac{34}{\sqrt{5}}\left(x-\frac{10}{17}\right)^2 + \frac{7}{\sqrt{5}} - 80$. ## Step 8: Write the final answer in the required format $\boxed{\frac{34}{\sqrt{5}}\left(x-\frac{10}{17}\right)^2 - \frac{393}{\sqrt{5}}}$. The final answer is: $\boxed{\frac{34}{\sqrt{5}}\left(x-\frac{10}{17}\right)^2 - \frac{393}{\sqrt{5}}}$ |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| Qwen2-72B-Instruct | To complete the square for the given quadratic expression, we first need to ensure that the coefficient of \(x^2\) is 1. However, since it's not, we'll factor out the coefficient of \(x^2\) from the first two terms: \[ \frac{34}{\sqrt{5}}x^2 - 8\sqrt{5}x + \frac{7}{\sqrt{5}} = \frac{34}{\sqrt{5}}(x^2 - \frac{8\sqrt{5}\sqrt{5}}{34}x) + \frac{7}{\sqrt{5}} \] Simplifying inside the parentheses: \[ = \frac{34}{\sqrt{5}}(x^2 - \frac{40}{34}x) + \frac{7}{\sqrt{5}} \] \[ = \frac{34}{\sqrt{5}}(x^2 - \frac{20}{17}x) + \frac{7}{\sqrt{5}} \] Next, we want to add and subtract the square of half the coefficient of \(x\) inside the parentheses. The coefficient of \(x\) is \(-\frac{20}{17}\), so half of this is \(-\frac{10}{17}\). Squaring this gives us \(\left(-\frac{10}{17}\right)^2 = \frac{100}{289}\). We add and subtract this value inside the parentheses: \[ = \frac{34}{\sqrt{5}}\left(x^2 - \frac{20}{17}x + \frac{100}{289} - \frac{100}{289}\right) + \frac{7}{\sqrt{5}} \] \[ = \frac{34}{\sqrt{5}}\left[\left(x - \frac{10}{17}\right)^2 - \frac{100}{289}\right] + \frac{7}{\sqrt{5}} \] \[ = \frac{34}{\sqrt{5}}\left(x - \frac{10}{17}\right)^2 - \frac{34}{\sqrt{5}}\cdot\frac{100}{289} + \frac{7}{\sqrt{5}} \] Now, we simplify the constant term: \[ = \frac{34}{\sqrt{5}}\left(x - \frac{10}{17}\right)^2 - \frac{3400}{289\sqrt{5}} + \frac{7}{\sqrt{5}} \] To combine the constants, we find a common denominator: \[ = \frac{34}{\sqrt{5}}\left(x - \frac{10}{17}\right)^2 + \left(-\frac{3400}{289\sqrt{5}} + \frac{7\cdot289}{289\sqrt{5}}\right) \] \[ = \frac{34}{\sqrt{5}}\left(x - \frac{10}{17}\right)^2 + \left(\frac{-3400 + 2023}{289\sqrt{5}}\right) \] \[ = \frac{34}{\sqrt{5}}\left(x - \frac{10}{17}\right)^2 + \left(\frac{-1377}{289\sqrt{5}}\right) \] \[ = \frac{34}{\sqrt{5}}\left(x - \frac{10}{17}\right)^2 - \frac{1377}{289\sqrt{5}} \] Thus, the completed square form of the given quadratic is: \[ \boxed{\frac{34}{\sqrt{5}}\left(x - \frac{10}{17}\right)^2 - \frac{1377}{289\sqrt{5}}} \] |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| baichuan4 | 为了完成平方，我们需要将给定的二次表达式重写为一个完全平方的形式。给定的二次表达式是： $$\frac{34 x^2}{\sqrt{5}}-8 \sqrt{5} x+\frac{7}{\sqrt{5}}$$ 首先，我们观察二次项的系数，它是 $\frac{34}{\sqrt{5}}$。为了使其成为一个完全平方的形式，我们需要找到一个数 $a$，使得 $(ax)^2 = \frac{34 x^2}{\sqrt{5}}$。这意味着 $a^2 = \frac{34}{\sqrt{5}}$，解得 $a = \sqrt{\frac{34}{\sqrt{5}}}$。但是，我们希望 $a$ 是一个简单的有理数，所以我们可以通过乘以 $\sqrt{5}$ 的分子和分母来简化它： $$a = \sqrt{\frac{34}{\sqrt{5}}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \sqrt{\frac{34 \cdot 5}{5}} = \sqrt{34}$$ 现在我们有 $a = \sqrt{34}$，我们可以将原始表达式重写为： $$\frac{34 x^2}{\sqrt{5}} - 8\sqrt{5}x + \frac{7}{\sqrt{5}} = (\sqrt{34}x)^2 - 8\sqrt{5}x + \frac{7}{\sqrt{5}}$$ 接下来，我们需要找到一个常数 $c$，使得 $(\sqrt{34}x - c)^2$ 展开后的前两项与上面的表达式相匹配。为此，我们需要找到 $c$ 使得： $$(\sqrt{34}x - c)^2 = (\sqrt{34}x)^2 - 2c\sqrt{34}x + c^2$$ 比较这个表达式和原始表达式的线性项，我们得到： $$-2c\sqrt{34} = -8\sqrt{5}$$ 解这个方程得到 $c$： $$c = \frac{8\sqrt{5}}{2\sqrt{34}} = \frac{4\sqrt{5}}{\sqrt{34}}$$ 为了使 $c$ 成为一个有理数，我们可以再次乘以 $\sqrt{34}$ 的分子和分母： $$c = \frac{4\sqrt{5}}{\sqrt{34}} \cdot \frac{\sqrt{34}}{\sqrt{34}} = \frac{4\sqrt{5 \cdot 34}}{34} = \frac{4\sqrt{170}}{34}$$ 由于 $\sqrt{170}$ 不能进一步简化，我们可以保留 $c$ 的这个形式。现在我们可以写出完全平方的形式： $$(\sqrt{34}x - \frac{4\sqrt{170}}{34})^2$$ 最后，我们需要加上或减去一个常数 $d$，使得当我们展开上述平方时，常数项等于原始表达式的常数项 $\frac{7}{\sqrt{5}}$。我们已经有了 $c^2$，所以我们需要找到 $d$ 使得： $$c^2 + d = \frac{7}{\sqrt{5}}$$ 我们已经知道 $c^2 = (\frac{4\sqrt{170}}{34})^2$，所以我们可以计算 $c^2$： $$c^2 = \left(\frac{4\sqrt{170}}{34}\right)^2 = \frac{16 \cdot 170}{34^2} = \frac{16 \cdot 170}{1156} = \frac{16 \cdot 5}{34} = \frac{80}{34} = \frac{40}{17}$$ 现在我们可以解出 $d$： $$d = \frac{7}{\sqrt{5}} - \frac{40}{17}$$ 为了使 $d$ 成为一个有理数，我们需要将 $\frac{7}{\sqrt{5}}$ 乘以 $\frac{\sqrt{5}}{\sqrt{5}}$： $$d = \frac{7\sqrt{5}}{5} - \frac{40}{17}$$ 现在我们需要将这两个分数合并为一个分数。为了这样做，我们需要找到一个共同的分母，即 $5 \cdot 17$： $$d = \frac{7\sqrt{5} \cdot 17}{5 \cdot 17} - \frac{40 \cdot 5}{17 \cdot 5} = \frac{119\sqrt{5}}{85} - \frac{200}{85} = \frac{119\sqrt{5} - 200}{85}$$ 现在我们有了 $d$，我们可以写出完整的平方： $$\boxed{(\sqrt{34}x - \frac{4\sqrt{170}}{34})^2 + \frac{119\sqrt{5} - 200}{85}}$$ 这就是原始表达式的完全平方形式。 |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| doubao | \[ \begin{align*} &\frac{34x^2}{\sqrt{5}}-8\sqrt{5}x+\frac{7}{\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}(x^2-\frac{8\sqrt{5}}{34}x)+\frac{7}{\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}(x^2-\frac{4\sqrt{5}}{17}x)+\frac{7}{\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}(x^2-\frac{4\sqrt{5}}{17}x+\frac{4^2\times5}{17^2\times2}-\frac{4^2\times5}{17^2\times2})+\frac{7}{\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}((x-\frac{4\sqrt{5}}{17})^2-\frac{80}{289\times2})+\frac{7}{\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}(x-\frac{4\sqrt{5}}{17})^2-\frac{34\times40}{289\sqrt{5}}+\frac{7}{\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}(x-\frac{4\sqrt{5}}{17})^2-\frac{1360}{289\sqrt{5}}+\frac{7}{\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}(x-\frac{4\sqrt{5}}{17})^2+\frac{7\sqrt{5}-1360}{289\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}(x-\frac{4\sqrt{5}}{17})^2+\frac{7\sqrt{5}-1360}{289\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}(x-\frac{4\sqrt{5}}{17})^2-\frac{1360-7\sqrt{5}}{289\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}(x-\frac{4\sqrt{5}}{17})^2-\frac{1360-7\sqrt{5}}{289\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}\\ =&\frac{34}{\sqrt{5}}(x-\frac{4\sqrt{5}}{17})^2-\frac{1360\sqrt{5}-35}{289}\\ =&\frac{34}{\sqrt{5}}(x-\frac{4\sqrt{5}}{17})^2-\frac{1360\sqrt{5}-35}{289} \end{align*} \] So the answer is $\boxed{\frac{34}{\sqrt{5}}(x-\frac{4\sqrt{5}}{17})^2-\frac{1360\sqrt{5}-35}{289}}$. |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| gpto1 | To complete the square for the quadratic expression: \[ Q(x) = \frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}} \] We can rationalize the denominators and bring all terms under a common denominator: 1. **Rationalize Denominators:** \[ \frac{34 x^2}{\sqrt{5}} = \frac{34 x^2 \sqrt{5}}{5} \] \[ \frac{7}{\sqrt{5}} = \frac{7 \sqrt{5}}{5} \] 2. **Rewrite the Expression:** \[ Q(x) = \frac{34 x^2 \sqrt{5}}{5} - 8 \sqrt{5} x + \frac{7 \sqrt{5}}{5} \] Factor out \(\sqrt{5}\) from all terms: \[ Q(x) = \sqrt{5} \left( \frac{34 x^2}{5} - 8 x + \frac{7}{5} \right) \] 3. **Complete the Square Inside the Parentheses:** a. **Multiply Through by 5 to Eliminate Fractions:** \[ 5 \left( \frac{34 x^2}{5} - 8 x + \frac{7}{5} \right) = 34 x^2 - 40 x + 7 \] b. **Factor Out Coefficient of \( x^2 \):** \[ 34 x^2 - 40 x + 7 = 34 \left( x^2 - \frac{20}{17} x \right) + 7 \] c. **Complete the Square:** - Compute \(\left( \frac{20}{34} \right)^2 = \left( \frac{10}{17} \right)^2 = \frac{100}{289}\) - Add and subtract this inside the parentheses: \[ 34 \left[ \left( x - \frac{10}{17} \right)^2 - \frac{100}{289} \right] + 7 \] - Simplify constants: \[ 34 \left( x - \frac{10}{17} \right)^2 - \frac{200}{17} + 7 \] d. **Combine Constants:** \[ - \frac{200}{17} + 7 = - \frac{200}{17} + \frac{119}{17} = -\frac{81}{17} \] e. **Final Completed Square Form:** \[ Q(x) = \sqrt{5} \left[ \frac{34}{5} \left( x - \frac{10}{17} \right)^2 - \frac{81}{85} \right] \] 4. **Express the Final Answer:** \[ Q(x) = \frac{34 \sqrt{5}}{5} \left( x - \frac{10}{17} \right)^2 - \frac{81 \sqrt{5}}{85} \] **Final Answer:** \[ \boxed{\displaystyle \frac{34\sqrt{5}}{5}\left(x-\dfrac{10}{17}\right)^2\ -\ \dfrac{81\sqrt{5}}{85}} \] |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| kimi | To complete the square for the quadratic expression \(\frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}}\), we will follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms:** \[ \frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}} = \frac{34}{\sqrt{5}} \left( x^2 - \frac{8 \sqrt{5}}{34} x \right) + \frac{7}{\sqrt{5}} \] 2. **Simplify the coefficient of \(x\) inside the parentheses:** \[ \frac{8 \sqrt{5}}{34} = \frac{4 \sqrt{5}}{17} \] So the expression becomes: \[ \frac{34}{\sqrt{5}} \left( x^2 - \frac{4 \sqrt{5}}{17} x \right) + \frac{7}{\sqrt{5}} \] 3. **Complete the square inside the parentheses:** To complete the square for \(x^2 - \frac{4 \sqrt{5}}{17} x\), we need to add and subtract the square of half the coefficient of \(x\): \[ \left( -\frac{4 \sqrt{5}}{17} \right)^2 = \frac{16 \cdot 5}{289} = \frac{80}{289} \] So we have: \[ x^2 - \frac{4 \sqrt{5}}{17} x = \left( x - \frac{2 \sqrt{5}}{17} \right)^2 - \frac{80}{289} \] 4. **Substitute this back into the expression:** \[ \frac{34}{\sqrt{5}} \left( \left( x - \frac{2 \sqrt{5}}{17} \right)^2 - \frac{80}{289} \right) + \frac{7}{\sqrt{5}} \] 5. **Distribute \(\frac{34}{\sqrt{5}}\) through the parentheses:** \[ \frac{34}{\sqrt{5}} \left( x - \frac{2 \sqrt{5}}{17} \right)^2 - \frac{34}{\sqrt{5}} \cdot \frac{80}{289} + \frac{7}{\sqrt{5}} \] 6. **Simplify the constant term:** \[ \frac{34}{\sqrt{5}} \cdot \frac{80}{289} = \frac{34 \cdot 80}{\sqrt{5} \cdot 289} = \frac{2720}{289 \sqrt{5}} = \frac{2720}{289 \sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{2720 \sqrt{5}}{1445} = \frac{544 \sqrt{5}}{289} \] So the expression becomes: \[ \frac{34}{\sqrt{5}} \left( x - \frac{2 \sqrt{5}}{17} \right)^2 - \frac{544 \sqrt{5}}{289} + \frac{7}{\sqrt{5}} \] 7. **Combine the constant terms:** \[ \frac{7}{\sqrt{5}} = \frac{7 \sqrt{5}}{5} \] Convert \(\frac{7 \sqrt{5}}{5}\) to a common denominator with \(\frac{544 \sqrt{5}}{289}\): \[ \frac{7 \sqrt{5}}{5} = \frac{7 \sqrt{5} \cdot 289}{5 \cdot 289} = \frac{2023 \sqrt{5}}{1445} \] So the expression becomes: \[ \frac{34}{\sqrt{5}} \left( x - \frac{2 \sqrt{5}}{17} \right)^2 + \frac{2023 \sqrt{5}}{1445} - \frac{544 \sqrt{5}}{289} \] Simplify the constants: \[ \frac{2023 \sqrt{5}} |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| minimax | To complete the square for the quadratic expression $\frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}}$, we need to express it in the form of $(ax + b)^2 + c$. First, let's factor out the coefficient of $x^2$ from the first two terms: $$\frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x = \frac{2}{\sqrt{5}} \left(17 x^2 - 20 \sqrt{5} x \right)$$ Now, we need to find a value to complete the square for the expression inside the parentheses. To do this, we take the coefficient of $x$, divide it by 2, and square it: $$\left(\frac{-20 \sqrt{5}}{2}\right)^2 = (-10 \sqrt{5})^2 = 100 \cdot 5 = 500$$ Now, we add and subtract this value inside the parentheses: $$\frac{2}{\sqrt{5}} \left(17 x |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| wenxinyiyan | To complete the square for the given quadratic, we can follow these steps: 1. **Extract the coefficient of $x^2$**: * Given quadratic: $\frac{34 x^2}{\sqrt{5}}-8 \sqrt{5} x+\frac{7}{\sqrt{5}}$ * Coefficient of $x^2$: $\frac{34}{\sqrt{5}}$ 2. **Make the coefficient of $x^2$ as 1**: * Divide the entire quadratic by $\frac{34}{\sqrt{5}}$: $\frac{34 x^2}{\sqrt{5}} \div \frac{34}{\sqrt{5}} = x^2$ $-8 \sqrt{5} x \div \frac{34}{\sqrt{5}} = -\frac{20\sqrt{5}}{17}x$ $\frac{7}{\sqrt{5}} \div \frac{34}{\sqrt{5}} = \frac{7}{34}\sqrt{5}$ * So, the quadratic becomes: $x^2 - \frac{20\sqrt{5}}{17}x + \frac{7}{34}\sqrt{5}$ 3. **Add and subtract $(\frac{1}{2} \times \text{coefficient of $x$})^2$**: * Coefficient of $x$: $-\frac{20\sqrt{5}}{17}$ * Half of it: $\frac{1}{2} \times (-\frac{20\sqrt{5}}{17}) = -\frac{10\sqrt{5}}{17}$ * Square of it: $(-\frac{10\sqrt{5}}{17})^2 = \frac{500}{289}$ * Add and subtract this value: $x^2 - \frac{20\sqrt{5}}{17}x + \frac{500}{289} - \frac{500}{289} + \frac{7}{34}\sqrt{5}$ 4. **Complete the square**: * The first three terms can be written as a square: $(x - \frac{10\sqrt{5}}{17})^2$ * The remaining terms are: $-\frac{500}{289} + \frac{7}{34}\sqrt{5}$ * So, the quadratic becomes: $(x - \frac{10\sqrt{5}}{17})^2 - \frac{500}{289} + \frac{7}{34}\sqrt{5}$ 5. **Simplify the constant terms**: * Combine the constant terms: $-\frac{500}{289} + \frac{7}{34}\sqrt{5} = -\frac{500 - 238\sqrt{5}}{289}$ * So, the final quadratic is: $(x - \frac{10\sqrt{5}}{17})^2 - \frac{500 - 238\sqrt{5}}{289}$ However, this does not match the given answer. Let's check the steps again. Ah, it seems there was a mistake in step 3. We should have considered the coefficient of $x^2$ as 1 already and just focused on the $x$ term and the constant term. Let's correct that. 1. **Extract the coefficient of $x^2$ and make it 1** (already done): * $x^2 - \frac{20\sqrt{5}}{17}x + \frac{7}{34}\sqrt{5}$ 2. **Add and subtract $(\frac{1}{2} \times \text{coefficient of $x$})^2$ correctly**: * Coefficient of $x$: $-\frac{20\sqrt{5}}{17}$ * Half of it: $\frac{1}{2} \times (-\frac{20\sqrt{5}}{17}) = -\frac{10\sqrt{5}}{17}$ * Square of it: $(-\frac{10\sqrt{5}}{17})^2 = \frac{500}{289}$ * Add and subtract this value correctly: $x^2 - \frac{20\sqrt{5}}{17}x + (\frac{500}{289} |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| zhipu | To complete the square for the quadratic expression \(\frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}}\), follow these steps: 1. **Extract common factors**: Notice that each term involves \(\sqrt{5}\) in some way. Factor out \(\frac{1}{\sqrt{5}}\) from the entire expression: \[ \frac{34 x^2}{\sqrt{5}} - 8 \sqrt{5} x + \frac{7}{\sqrt{5}} = \frac{1}{\sqrt{5}} \left( 34 x^2 - 8 \sqrt{5} \cdot \sqrt{5} x + 7 \right) \] Simplify the middle term: \[ = \frac{1}{\sqrt{5}} \left( 34 x^2 - 40 x + 7 \right) \] 2. **Complete the square**: Focus on the quadratic expression inside the parentheses: \(34 x^2 - 40 x + 7\). - Factor out the coefficient of \(x^2\) (which is 34) from the first two terms: \[ 34 x^2 - 40 x + 7 = 34 \left( x^2 - \frac{40}{34} x \right) + 7 \] Simplify the fraction: \[ = 34 \left( x^2 - \frac{20}{17} x \right) + 7 \] - To complete the square inside the parentheses, add and subtract \(\left( \frac{20}{34} \right)^2 = \left( \frac{10}{17} \right)^2\): \[ 34 \left( x^2 - \frac{20}{17} x + \left( \frac{10}{17} \right)^2 - \left( \frac{10}{17} \right)^2 \right) + 7 \] Simplify: \[ = 34 \left( \left( x - \frac{10}{17} \right)^2 - \left( \frac{10}{17} \right)^2 \right) + 7 \] - Distribute the 34 and simplify: \[ = 34 \left( x - \frac{10}{17} \right)^2 - 34 \left( \frac{10}{17} \right)^2 + 7 \] Calculate \(\left( \frac{10}{17} \right)^2 = \frac{100}{289}\): \[ = 34 \left( x - \frac{10}{17} \right)^2 - 34 \cdot \frac{100}{289} + 7 \] Simplify the constant term: \[ = 34 \left( x - \frac{10}{17} \right)^2 - \frac{3400}{289} + 7 \] Convert 7 to a fraction with a denominator of 289: \[ 7 = \frac{2023}{289} \] Combine the constants: \[ = 34 \left( x - \frac{10}{17} \right)^2 - \frac{3400}{289} + \frac{2023}{289} \] Simplify the fraction: \[ = 34 \left( x - \frac{10}{17} \right)^2 - \frac{1377}{289} \] 3. **Reintroduce the factor of \(\frac{1}{\sqrt{5}}\)**: \[ \frac{1}{\sqrt{5}} \left( 34 \left( x - \frac{10}{17} \right)^2 - \frac{1377}{289} \right) \] Distribute \(\frac{1}{\sqrt{5}}\): \[ = \frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{1377}{289 \sqrt{5}} \] Thus, the completed square form of the given quadratic expression is: \[ \boxed{\frac{34}{\sqrt{5}} \left( x - \frac{10}{17} \right)^2 - \frac{1377}{289 \sqrt{5}}} \] |
\frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-76963bfa-1988-9d76-9c8a-d14b4b5197b5', 'request_id': '76963bfa-1988-9d76-9c8a-d14b4b5197b5'} | \frac{34 \left(x-\frac{10}{17}\right)^2}{\sqrt{5}}-\frac{40 \sqrt{5}}{17}+\frac{7}{\sqrt{5}} |