| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To complete the square for the quadratic expression \(6x^2 - 4x + 2\), we will follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms:** \[ 6x^2 - 4x + 2 = 6(x^2 - \frac{2}{3}x) + 2 \] 2. **Complete the square inside the parentheses:** - Take the coefficient of \(x\) inside the parentheses, which is \(-\frac{2}{3}\), divide it by 2, and square it: \[ \left(\frac{-\frac{2}{3}}{2}\right)^2 = \left(\frac{-1}{3}\right)^2 = \frac{1}{9} \] - Add and subtract this square inside the parentheses: \[ 6\left(x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}\right) + 2 \] - This can be rewritten as: \[ 6\left((x - \frac{1}{3})^2 - \frac{1}{9}\right) + 2 \] 3. **Distribute the 6 and simplify:** \[ 6(x - \frac{1}{3})^2 - 6 \cdot \frac{1}{9} + 2 = 6(x - \frac{1}{3})^2 - \frac{2}{3} + 2 \] - Combine the constants: \[ -\frac{2}{3} + 2 = -\frac{2}{3} + \frac{6}{3} = \frac{4}{3} \] 4. **Write the final expression:** \[ 6(x - \frac{1}{3})^2 + \frac{4}{3} \] Thus, the completed square form of the quadratic expression \(6x^2 - 4x + 2\) is: \[ \boxed{6(x - \frac{1}{3})^2 + \frac{4}{3}} \] |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| deepseek-r1 | To complete the square for the quadratic \(6x^2 - 4x + 2\): 1. **Factor out the coefficient of \(x^2\) from the first two terms**: \[ 6(x^2 - \frac{2}{3}x) + 2 \] 2. **Complete the square inside the parentheses**: - The coefficient of \(x\) is \(-\frac{2}{3}\). Half of this coefficient is \(-\frac{1}{3}\), and squaring it gives \(\left(-\frac{1}{3}\right)^2 = \frac{1}{9}\). - Add and subtract \(\frac{1}{9}\) inside the parentheses: \[ 6\left(x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}\right) + 2 \] 3. **Rewrite the expression**: - The quadratic expression inside the parentheses becomes a perfect square: \[ 6\left(\left(x - \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 2 \] 4. **Distribute and simplify**: - Distribute the 6 and simplify the constants: \[ 6\left(x - \frac{1}{3}\right)^2 - 6 \cdot \frac{1}{9} + 2 \] \[ 6\left(x - \frac{1}{3}\right)^2 - \frac{2}{3} + 2 \] \[ 6\left(x - \frac{1}{3}\right)^2 + \frac{4}{3} \] The completed square form of the quadratic \(6x^2 - 4x + 2\) is \(\boxed{6\left(x - \dfrac{1}{3}\right)^2 + \dfrac{4}{3}}\). |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| gpto1 | To complete the square for the quadratic \(6x^2 - 4x + 2\), we follow these steps: 1. Factor out the coefficient of the \(x^2\) term from the first two terms: \[ 6x^2 - 4x + 2 = 6\left( x^2 - \frac{2}{3}x \right) + 2 \] 2. Complete the square inside the parentheses. Compute \(\left(\frac{\text{coefficient of }x}{2}\right)^2\): \[ \left( \frac{-\frac{2}{3}}{2} \right)^2 = \left( -\frac{1}{3} \right)^2 = \frac{1}{9} \] 3. Add and subtract this value inside the parentheses: \[ 6\left( x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9} \right) + 2 = 6\left( \left( x - \frac{1}{3} \right)^2 - \frac{1}{9} \right) + 2 \] 4. Distribute the 6 and simplify: \[ 6\left( x - \frac{1}{3} \right)^2 - \frac{2}{3} + 2 = 6\left( x - \frac{1}{3} \right)^2 + \frac{4}{3} \] So the completed square form of the quadratic is: \[ \boxed{6\left( x - \dfrac{1}{3} \right)^2 + \dfrac{4}{3}} \] \(\boxed{6\left( x-\dfrac{1}{3} \right)^{2}\,+\,\dfrac{4}{3}}\) |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| gpt-4o-2024-08-06 | To complete the square for the quadratic expression \(6x^2 - 4x + 2\), we follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms:** The coefficient of \(x^2\) is 6. We factor 6 out of the first two terms: \[ 6x^2 - 4x = 6(x^2 - \frac{2}{3}x) \] 2. **Complete the square inside the parentheses:** To complete the square for the expression \(x^2 - \frac{2}{3}x\), we need to find a constant that makes it a perfect square trinomial. Take half of the coefficient of \(x\), which is \(-\frac{2}{3}\), and square it: \[ \left(-\frac{2}{3} \times \frac{1}{2}\right)^2 = \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \] Add and subtract \(\frac{1}{9}\) inside the parentheses: \[ x^2 - \frac{2}{3}x = \left(x^2 - \frac{2}{3}x + \frac{1}{9}\right) - \frac{1}{9} \] The expression inside the parentheses is now a perfect square: \[ x^2 - \frac{2}{3}x + \frac{1}{9} = \left(x - \frac{1}{3}\right)^2 \] 3. **Substitute back and simplify:** Substitute back into the expression: \[ 6(x^2 - \frac{2}{3}x) = 6\left(\left(x - \frac{1}{3}\right)^2 - \frac{1}{9}\right) \] Distribute the 6: \[ 6\left(x - \frac{1}{3}\right)^2 - 6 \times \frac{1}{9} = 6\left(x - \frac{1}{3}\right)^2 - \frac{2}{3} \] 4. **Add the constant term from the original expression:** The original expression is \(6x^2 - 4x + 2\). We have: \[ 6\left(x - \frac{1}{3}\right)^2 - \frac{2}{3} + 2 \] Combine the constant terms: \[ -\frac{2}{3} + 2 = -\frac{2}{3} + \frac{6}{3} = \frac{4}{3} \] So the completed square form is: \[ 6\left(x - \frac{1}{3}\right)^2 + \frac{4}{3} \] Thus, the completed square form of the quadratic is: \[ \boxed{6\left(x - \frac{1}{3}\right)^2 + \frac{4}{3}} \] |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| DeepSeek-Coder-V2-Lite-Instruct | To complete the square for the quadratic expression \(6x^2 - 4x + 2\), we will follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms:** \[ 6x^2 - 4x + 2 = 6(x^2 - \frac{2}{3}x) + 2 \] 2. **Complete the square inside the parentheses:** - Take the coefficient of \(x\) inside the parentheses, which is \(-\frac{2}{3}\). - Divide it by 2: \(\frac{-\frac{2}{3}}{2} = -\frac{1}{3}\). - Square this result: \(\left(-\frac{1}{3}\right)^2 = \frac{1}{9}\). 3. **Add and subtract this square inside the parentheses:** \[ 6\left(x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}\right) + 2 \] \[ = 6\left(\left(x - \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 2 \] 4. **Distribute the 6 and simplify:** \[ = 6\left(x - \frac{1}{3}\right)^2 - 6 \cdot \frac{1}{9} + 2 \] \[ = 6\left(x - \frac{1}{3}\right)^2 - \frac{6}{9} + 2 \] \[ = 6\left(x - \frac{1}{3}\right)^2 - \frac{2}{3} + 2 \] \[ = 6\left(x - \frac{1}{3}\right)^2 + \frac{4}{3} \] Thus, the completed square form of the quadratic expression \(6x^2 - 4x + 2\) is: \[ \boxed{6\left(x - \frac{1}{3}\right)^2 + \frac{4}{3}} \] |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the coefficients of the quadratic equation The given quadratic equation is $6x^2 - 4x + 2$. The coefficients are $a = 6$, $b = -4$, and $c = 2$. ## Step 2: Divide the entire equation by the coefficient of $x^2$ to make the coefficient of $x^2$ equal to 1 To simplify the process, divide the entire equation by $a = 6$. This gives us $x^2 - \frac{2}{3}x + \frac{1}{3}$. ## Step 3: Move the constant term to the other side of the equation Since we are completing the square, we need to isolate the $x^2$ and $x$ terms. So, we move the constant term $\frac{1}{3}$ to the other side, resulting in $x^2 - \frac{2}{3}x = -\frac{1}{3}$. ## Step 4: Find the value to complete the square To complete the square, we take half of the coefficient of $x$, square it, and add it to both sides. Half of $-\frac{2}{3}$ is $-\frac{1}{3}$, and squaring it gives $\frac{1}{9}$. ## Step 5: Add the value to complete the square to both sides of the equation Adding $\frac{1}{9}$ to both sides gives us $x^2 - \frac{2}{3}x + \frac{1}{9} = -\frac{1}{3} + \frac{1}{9}$. ## Step 6: Simplify the right side of the equation Combining the fractions on the right side gives us $-\frac{3}{9} + \frac{1}{9} = -\frac{2}{9}$. ## Step 7: Write the completed square The completed square form is $\left(x - \frac{1}{3}\right)^2 = -\frac{2}{9}$. ## Step 8: Multiply both sides by the original coefficient of $x^2$ to get the final form Multiplying both sides by $6$ gives $6\left(x - \frac{1}{3}\right)^2 = -\frac{4}{3}$. ## Step 9: Write the final form of the completed square The final form of the completed square for the given quadratic is $6\left(x - \frac{1}{3}\right)^2 - \frac{4}{3}$. ## Step 10: Add the constant term from the original equation However, we need to remember that we originally moved the constant term $\frac{1}{3}$ to the other side. To get the correct form, we need to add $2$ (the original constant term) inside the expression. ## Step 11: Write the correct form of the completed square However, from the previous steps we can tell that we can't just simply add 2. So we just simplify the expression from step 9. The final answer is: $\boxed{6\left(x - \frac{1}{3}\right)^2 - \frac{4}{3}}$ |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the coefficients of the quadratic equation The given quadratic equation is $6x^2 - 4x + 2$. The coefficients are $a = 6$, $b = -4$, and $c = 2$. ## Step 2: Determine the coefficient of the squared term The coefficient of the squared term is $a = 6$. To complete the square, we need to factor this coefficient out of the first two terms. ## Step 3: Factor out the coefficient of the squared term Factoring out $6$ from the first two terms gives $6(x^2 - \frac{2}{3}x) + 2$. ## Step 4: Calculate the value to complete the square To complete the square, we need to add and subtract $(\frac{b}{2})^2$ inside the parentheses. In this case, $\frac{b}{2} = \frac{-2/3}{2} = -\frac{1}{3}$. So, $(\frac{b}{2})^2 = (-\frac{1}{3})^2 = \frac{1}{9}$. ## Step 5: Add and subtract the calculated value inside the parentheses Adding and subtracting $\frac{1}{9}$ inside the parentheses gives $6(x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}) + 2$. ## Step 6: Rewrite the expression with the completed square Rewriting the expression gives $6((x - \frac{1}{3})^2 - \frac{1}{9}) + 2$. ## Step 7: Simplify the expression Simplifying the expression gives $6(x - \frac{1}{3})^2 - \frac{6}{9} + 2$. ## Step 8: Combine like terms Combining like terms gives $6(x - \frac{1}{3})^2 - \frac{2}{3} + 2$. ## Step 9: Simplify further Simplifying further gives $6(x - \frac{1}{3})^2 + \frac{4}{3}$. ## Step 10: Write the final answer in the required format The final answer is: $\boxed{6(x - \frac{1}{3})^2 + \frac{4}{3}}$ |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| Qwen2-72B-Instruct | To complete the square for the quadratic expression $6x^2 - 4x + 2$, we first factor out the coefficient of $x^2$ from the first two terms: \[6x^2 - 4x + 2 = 6(x^2 - \frac{4}{6}x) + 2\] Simplify the fraction inside the parentheses: \[= 6(x^2 - \frac{2}{3}x) + 2\] Next, to complete the square, we need to add and subtract the square of half the coefficient of $x$ inside the parentheses. The coefficient of $x$ is $-\frac{2}{3}$, so half of that is $-\frac{1}{3}$, and its square is $\left(-\frac{1}{3}\right)^2 = \frac{1}{9}$. We multiply this by the factor outside the parentheses (which is 6) to ensure we're adding and subtracting the same value: \[= 6\left(x^2 - \frac{2}{3}x + \left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right)^2\right) + 2\] \[= 6\left(x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}\right) + 2\] \[= 6\left(\left(x - \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 2\] \[= 6\left(x - \frac{1}{3}\right)^2 - 6\cdot\frac{1}{9} + 2\] \[= 6\left(x - \frac{1}{3}\right)^2 - \frac{2}{3} + 2\] \[= 6\left(x - \frac{1}{3}\right)^2 + \frac{4}{3}\] Thus, the completed square form of $6x^2 - 4x + 2$ is: \[\boxed{6\left(x - \frac{1}{3}\right)^2 + \frac{4}{3}}\] |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| baichuan4 | To complete the square for the quadratic $6x^2 - 4x + 2$, we first need to identify the coefficients of the quadratic terms. We have $a = 6$, $b = -4$, and $c = 2$. The goal is to rewrite the quadratic in the form $(ax^2 + bx + c) = (a)(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola represented by the quadratic equation. Step 1: Divide the entire equation by the coefficient of $x^2$, which is $a = 6$. This gives us: $$ x^2 - \frac{2}{3}x + \frac{1}{3} $$ Step 2: Complete the square. To do this, we take half of the coefficient of $x$, square it, and add it to both sides of the equation. The coefficient of $x$ is $-\frac{2}{3}$, so half of it is $-\frac{1}{3}$. Squaring this gives us $\left(-\frac{1}{3}\right)^2 = \frac{1}{9}$. Step 3: Add $\frac{1}{9}$ to both sides of the equation: $$ x^2 - \frac{2}{3}x + \frac{1}{9} = \frac{1}{3} + \frac{1}{9} $$ Step 4: Simplify the right side of the equation. The sum of $\frac{1}{3}$ and $\frac{1}{9}$ is $\frac{3}{9} + \frac{1}{9} = \frac{4}{9}$. Step 5: Write the left side of the equation as a perfect square. We have $x^2 - \frac{2}{3}x + \frac{1}{9}$, which is the same as $(x - \frac{1}{3})^2$. Step 6: Our final equation is: $$ (x - \frac{1}{3})^2 = \frac{4}{9} $$ Step 7: Multiply both sides of the equation by $6$ to bring it back to its original form: $$ 6(x - \frac{1}{3})^2 = 6 \cdot \frac{4}{9} $$ Step 8: Simplify the right side of the equation. The product of $6$ and $\frac{4}{9}$ is $\frac{24}{9}$, which simplifies to $\frac{8}{3}$. Step 9: Put the final answer in a box: $$ \boxed{6(x - \frac{1}{3})^2 + \frac{8}{3}} $$ This is the completed square form of the original quadratic equation $6x^2 - 4x + 2$. |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| doubao | $6x^2 - 4x + 2 = 6(x^2 - \frac{2}{3}x) + 2$ $= 6(x^2 - \frac{2}{3}x + (\frac{1}{3})^2 - (\frac{1}{3})^2) + 2$ $= 6((x - \frac{1}{3})^2 - \frac{1}{9}) + 2$ $= 6(x - \frac{1}{3})^2 - \frac{2}{3} + 2$ $= 6(x - \frac{1}{3})^2 + \frac{4}{3}$ $=\boxed{6(x - \frac{1}{3})^2 + \frac{4}{3}}$. |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| gpto1 | To complete the square for the quadratic \(6x^2 - 4x + 2\), we follow these steps: 1. Factor out the coefficient of the \(x^2\) term from the first two terms: \[ 6x^2 - 4x + 2 = 6\left( x^2 - \frac{2}{3}x \right) + 2 \] 2. Complete the square inside the parentheses. Compute \(\left(\frac{\text{coefficient of }x}{2}\right)^2\): \[ \left( \frac{-\frac{2}{3}}{2} \right)^2 = \left( -\frac{1}{3} \right)^2 = \frac{1}{9} \] 3. Add and subtract this value inside the parentheses: \[ 6\left( x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9} \right) + 2 = 6\left( \left( x - \frac{1}{3} \right)^2 - \frac{1}{9} \right) + 2 \] 4. Distribute the 6 and simplify: \[ 6\left( x - \frac{1}{3} \right)^2 - \frac{2}{3} + 2 = 6\left( x - \frac{1}{3} \right)^2 + \frac{4}{3} \] So the completed square form of the quadratic is: \[ \boxed{6\left( x - \dfrac{1}{3} \right)^2 + \dfrac{4}{3}} \] \(\boxed{6\left( x-\dfrac{1}{3} \right)^{2}\,+\,\dfrac{4}{3}}\) |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| kimi | To complete the square for the quadratic expression \(6x^2 - 4x + 2\), we will follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms:** \[ 6x^2 - 4x + 2 = 6(x^2 - \frac{2}{3}x) + 2 \] 2. **Identify the coefficient of \(x\) inside the parentheses:** The coefficient is \(-\frac{2}{3}\). 3. **Determine the value to complete the square:** To complete the square, we need to add and subtract the square of half the coefficient of \(x\). First, find half of \(-\frac{2}{3}\): \[ \frac{-\frac{2}{3}}{2} = -\frac{1}{3} \] Then square this value: \[ \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \] 4. **Add and subtract this square inside the parentheses:** \[ 6\left(x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}\right) + 2 \] 5. **Rewrite the expression inside the parentheses as a perfect square trinomial:** \[ 6\left((x - \frac{1}{3})^2 - \frac{1}{9}\right) + 2 \] 6. **Distribute the 6 through the parentheses:** \[ 6(x - \frac{1}{3})^2 - 6 \cdot \frac{1}{9} + 2 \] Simplify the multiplication: \[ 6(x - \frac{1}{3})^2 - \frac{6}{9} + 2 \] Simplify \(\frac{6}{9}\): \[ 6(x - \frac{1}{3})^2 - \frac{2}{3} + 2 \] 7. **Convert 2 to a fraction with a common denominator:** \[ 2 = \frac{6}{3} \] So the expression becomes: \[ 6(x - \frac{1}{3})^2 - \frac{2}{3} + \frac{6}{3} \] 8. **Combine the constant terms:** \[ 6(x - \frac{1}{3})^2 + \frac{4}{3} \] Thus, the completed square form of the quadratic expression \(6x^2 - 4x + 2\) is: \[ \boxed{6(x - \frac{1}{3})^2 + \frac{4}{3}} \] |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| minimax | To complete the square for the quadratic $6x^2 - 4x + 2$, we need to express it in the form $(ax + b)^2 + c$. Here are the steps: 1. Factor out the coefficient of $x^2$ from the first two terms: $$6(x^2 - \frac{2}{3}x) + 2$$ 2. Find the value that completes the square for the expression inside the parentheses. To do this, take half of the coefficient of $x$ and square it. In this case, half of $-\frac{2}{3}$ is $-\frac{1}{3}$, and squaring $-\frac{1}{3}$ gives us $\left(-\frac{1}{3}\right)^2 = \frac{1}{9}$. 3. Add and subtract this value inside the parentheses to complete the square: $$6\left(x^2 - \frac{2}{3}x + \frac{1}{9}\right) - 6\left(\frac{1}{9}\right) + 2$$ 4. Simplify the expression: $$6\left(x - \frac{1}{3}\right)^2 - |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| wenxinyiyan | 为了完成平方，我们首先需要找到$a$和$b$的值，其中$a$是二次项系数，$b$是一次项系数。在这个问题中，$a = 6$，$b = -4$。 完成平方的一般形式是$(x - h)^{2} + k$。为了将原二次式转化为这种形式，我们需要找到一个数$h$，使得$x^2 - \frac{b}{a}x$可以写成$(x - h)^{2}$的形式。 步骤1：找到$h$的值。 $h = \frac{b}{2a} = \frac{-4}{2 \times 6} = -\frac{1}{3}$ 步骤2：将原二次式重写为完全平方的形式。 $6x^{2} - 4x + 2 = 6(x^{2} - \frac{2}{3}x) + 2$ $= 6(x^{2} - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}) + 2$ $= 6((x - \frac{1}{3})^{2} - \frac{1}{9}) + 2$ $= 6(x - \frac{1}{3})^{2} - \frac{2}{3} + 2$ $= 6(x - \frac{1}{3})^{2} + \frac{4}{3}$ 所以，完成平方后的二次式为： $\boxed{6(x - \frac{1}{3})^{2} + \frac{4}{3}}$ |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| zhipu | To complete the square for the quadratic $6x^2 - 4x + 2$, we follow these steps: 1. **Factor out the leading coefficient** from the terms involving $x^2$ and $x$: \[ 6x^2 - 4x + 2 = 6(x^2 - \frac{2}{3}x) + 2 \] 2. **Complete the square** inside the parentheses: - Take half of the coefficient of $x$, which is $-\frac{2}{3}$, and square it: \[ \left(-\frac{2}{3} \cdot \frac{1}{2}\right)^2 = \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \] - Add and subtract this square inside the parentheses: \[ 6\left(x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}\right) + 2 \] - Simplify the expression inside the parentheses: \[ 6\left((x - \frac{1}{3})^2 - \frac{1}{9}\right) + 2 \] 3. **Distribute the 6** and combine the constants: - Distribute the 6 to both terms inside the parentheses: \[ 6(x - \frac{1}{3})^2 - 6 \cdot \frac{1}{9} + 2 \] - Simplify the constants: \[ 6(x - \frac{1}{3})^2 - \frac{6}{9} + 2 = 6(x - \frac{1}{3})^2 - \frac{2}{3} + 2 \] - Combine the constants: \[ 6(x - \frac{1}{3})^2 + 2 - \frac{2}{3} = 6(x - \frac{1}{3})^2 + \frac{6}{3} - \frac{2}{3} = 6(x - \frac{1}{3})^2 + \frac{4}{3} \] Thus, the completed square form of the quadratic $6x^2 - 4x + 2$ is: \[ \boxed{6(x - \frac{1}{3})^2 + \frac{4}{3}} \] |
6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-0d89866d-0b3f-9688-92e8-04441ee61898', 'request_id': '0d89866d-0b3f-9688-92e8-04441ee61898'} | 6 \left(x-\frac{1}{3}\right)^2+\frac{4}{3} |