| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To simplify the fraction, we need to rationalize the denominator. We can do this by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of $\sqrt{108}+2\sqrt{12}+2\sqrt{27}$ is $\sqrt{108}-2\sqrt{12}-2\sqrt{27}$. So, we multiply the numerator and denominator by $\sqrt{108}-2\sqrt{12}-2\sqrt{27}$: $$\frac{4}{\sqrt{108}+2\sqrt{12}+2\sqrt{27}} \cdot \frac{\sqrt{108}-2\sqrt{12}-2\sqrt{27}}{\sqrt{108}-2\sqrt{12}-2\sqrt{27}}$$ This simplifies to: $$\frac{4(\sqrt{108}-2\sqrt{12}-2\sqrt{27})}{(\sqrt{108}+2\sqrt{12}+2\sqrt{27})(\sqrt{108}-2\sqrt{12}-2\sqrt{27})}$$ Now, we can simplify the denominator using the difference of squares formula: $$(\sqrt{108}+2\sqrt{12}+2\sqrt{27})(\sqrt{108}-2\sqrt{12}-2\sqrt{27}) = (\sqrt{108})^2 - (2\sqrt{12}+2\sqrt{27})^2$$ $$= 108 - (4\sqrt{12}+4\sqrt{27}+4\sqrt{12}\sqrt{27})$$ $$= 108 - 4(\sqrt{12}+\sqrt{27}+\sqrt{12}\sqrt{27})$$ $$= 108 - 4(2\sqrt{3}+3\sqrt{3}+6)$$ $$= 108 - 4(5\sqrt{3}+6)$$ $$= 108 - 20\sqrt{3} - 24$$ $$= 84 - 20\sqrt{3}$$ Now, we can simplify the numerator: $$4(\sqrt{108}-2\sqrt{12}-2\sqrt{27}) = 4(6\sqrt{3}-4\sqrt{3}-6\sqrt{3})$$ $$= 4(-4\sqrt{3})$$ $$= -16\sqrt{3}$$ So, the simplified fraction is: $$\frac{-16\sqrt{3}}{84-20\sqrt{3}}$$ We can further simplify this fraction by dividing both the numerator and denominator by $-4\sqrt{3}$: $$\frac{-16\sqrt{3}}{84-20\sqrt{3}} \cdot \frac{-4\sqrt{3}}{-4\sqrt{3}}$$ This simplifies to: $$\frac{4\sqrt{3}}{21-5\sqrt{3}}$$ Therefore, the correct answer is: A. $\frac{4\sqrt{3}}{21-5\sqrt{3}}$ However, this is not among the answer choices. It seems that there was an error in the simplification process. Let's re-evaluate the expression. $$\frac{4}{\sqrt{108}+2\sqrt{12}+2\sqrt{27}}$$ We can simplify the denominator by combining like terms: $$\sqrt{108}+2\sqrt{12}+2\sqrt{27} = \sqrt{36 \cdot 3} + 2\sqrt{4 \cdot 3} + 2\sqrt{9 \cdot 3}$$ $$= 6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3}$$ $$= 16\sqrt{3}$$ So, the fraction becomes: $$\frac{4}{16\sqrt{3}}$$ We can simplify this fraction by dividing both the numerator and denominator by 4: $$\frac{4}{16\sqrt{3}} \cdot \frac{4}{4}$$ This simplifies to: $$\frac{1}{4\sqrt{3}}$ Therefore, the correct answer is: D. $\frac{1}{4\sqrt{3}}$. The best answer is D |
B |
| deepseekr1 | To simplify the given fraction by rationalizing the denominator, we first simplify each term in the denominator: 1. **Simplify √108**: \( \sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3} \). 2. **Simplify \( 2\sqrt{12} \):** \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \), so \( 2\sqrt{12} = 2 \times 2\sqrt{3} = 4\sqrt{3} \). 3. **Simplify \( 2\sqrt{27} \):** \( \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \), so \( 2\sqrt{27} = 2 \times 3\sqrt{3} = 6\sqrt{3} \). **Combine the simplified terms in the denominator:** \( 6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3} = (6 + 4 + 6)\sqrt{3} = 16\sqrt{3} \). **Substitute back into the original fraction:** \( \frac{4}{16\sqrt{3}} = \frac{1}{4\sqrt{3}} \). **Rationalize the denominator:** Multiply numerator and denominator by \( \sqrt{3} \): \( \frac{1 \times \sqrt{3}}{4\sqrt{3} \times \sqrt{3}} = \frac{\sqrt{3}}{4 \times 3} = \frac{\sqrt{3}}{12} \). **Answer:** The simplified fraction is \( \frac{\sqrt{3}}{12} \), which corresponds to **option B**. **Answer:** B |
B |
| gpto1 | To solve this problem, we'll simplify each radical and then rationalize the denominator. 1. **Simplify the radicals:** - \(\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\) - \(2\sqrt{12} = 2 \times \sqrt{4 \times 3} = 2 \times 2\sqrt{3} = 4\sqrt{3}\) - \(2\sqrt{27} = 2 \times \sqrt{9 \times 3} = 2 \times 3\sqrt{3} = 6\sqrt{3}\) 2. **Combine like terms in the denominator:** \(6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3} = 16\sqrt{3}\) 3. **Simplify the fraction:** \(\frac{4}{16\sqrt{3}} = \frac{1}{4\sqrt{3}}\) 4. **Rationalize the denominator:** \(\frac{1}{4\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{12}\) So, the simplified fraction is \(\frac{\sqrt{3}}{12}\), which corresponds to option **B**. **Answer: B** |
B |
| GPT-4o | To simplify the fraction \(\frac{4}{\sqrt{108} + 2\sqrt{12} + 2\sqrt{27}}\), we first need to simplify the terms in the denominator. 1. Simplify \(\sqrt{108}\): \[ \sqrt{108} = \sqrt{36 \times 3} = \sqrt{36} \times \sqrt{3} = 6\sqrt{3} \] 2. Simplify \(2\sqrt{12}\): \[ 2\sqrt{12} = 2 \times \sqrt{4 \times 3} = 2 \times \sqrt{4} \times \sqrt{3} = 2 \times 2 \times \sqrt{3} = 4\sqrt{3} \] 3. Simplify \(2\sqrt{27}\): \[ 2\sqrt{27} = 2 \times \sqrt{9 \times 3} = 2 \times \sqrt{9} \times \sqrt{3} = 2 \times 3 \times \sqrt{3} = 6\sqrt{3} \] Now, combine these simplified terms in the denominator: \[ \sqrt{108} + 2\sqrt{12} + 2\sqrt{27} = 6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3} = (6 + 4 + 6)\sqrt{3} = 16\sqrt{3} \] So, the original fraction simplifies to: \[ \frac{4}{\sqrt{108} + 2\sqrt{12} + 2\sqrt{27}} = \frac{4}{16\sqrt{3}} \] Next, simplify the fraction: \[ \frac{4}{16\sqrt{3}} = \frac{4}{16} \times \frac{1}{\sqrt{3}} = \frac{1}{4} \times \frac{1}{\sqrt{3}} = \frac{1}{4\sqrt{3}} \] Now, rationalize the denominator by multiplying the numerator and the denominator by \(\sqrt{3}\): \[ \frac{1}{4\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{4 \times 3} = \frac{\sqrt{3}}{12} \] Therefore, the correct answer is: B. \(\frac{\sqrt{3}}{12}\) |
B |
| llama3.1-70B-Instruct | A First, we simplify the denominator by factoring out the perfect squares: \begin{align*} \sqrt{108}+2\sqrt{12}+2\sqrt{27}&=\sqrt{36\cdot3}+2\sqrt{4\cdot3}+2\sqrt{9\cdot3}\\ &=6\sqrt{3}+4\sqrt{3}+6\sqrt{3}\\ &=16\sqrt{3}. \end{align*}Then, we can simplify the fraction by rationalizing the denominator: \begin{align*} \frac{4}{\sqrt{108}+2\sqrt{12}+2\sqrt{27}}&=\frac{4}{16\sqrt{3}}\\ &=\frac{4}{16\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\\ &=\frac{4\sqrt{3}}{16\cdot3}\\ &=\boxed{\frac{\sqrt{3}}{12}}. \end{align*} Final Answer: The final answer is B. I hope it is correct. |
B |
| llama3.1-405B-Instruct | A The best answer is A |
B |
| DeepSeek-Coder-V2 | To simplify the fraction by rationalizing the denominator, we start with the given expression: \[ \frac{4}{\sqrt{108} + 2\sqrt{12} + 2\sqrt{27}} \] First, we simplify each term in the denominator: \[ \sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3} \] \[ 2\sqrt{12} = 2\sqrt{4 \cdot 3} = 2 \cdot 2\sqrt{3} = 4\sqrt{3} \] \[ 2\sqrt{27} = 2\sqrt{9 \cdot 3} = 2 \cdot 3\sqrt{3} = 6\sqrt{3} \] Now, substitute these simplified terms back into the denominator: \[ \sqrt{108} + 2\sqrt{12} + 2\sqrt{27} = 6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3} = 16\sqrt{3} \] So the expression becomes: \[ \frac{4}{16\sqrt{3}} \] Simplify the fraction: \[ \frac{4}{16\sqrt{3}} = \frac{1}{4\sqrt{3}} \] Thus, the simplified form of the given expression is: \[ \boxed{D} \] |
B |
| Qwen2-72B-Instruct | A. \frac{4\sqrt{3}}{33} | B |
| baichuan4 | To rationalize the denominator of the fraction $$\frac{4}{\sqrt{108}+2\sqrt{12}+2\sqrt{27}}$$, we need to multiply it by a form of 1 that will eliminate the radicals in the denominator. We can do this by multiplying the numerator and denominator by the conjugate of the denominator, which is $$\sqrt{108}-2\sqrt{12}-2\sqrt{27}$$. Let's simplify the denominators first: - $$\sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3}$$ - $$2\sqrt{12} = 2\sqrt{4 \cdot 3} = 4\sqrt{3}$$ - $$2\sqrt{27} = 2\sqrt{9 \cdot 3} = 6\sqrt{3}$$ Substituting these back into the original fraction, we get: $$\frac{4}{6\sqrt{3}+4\sqrt{3}+6\sqrt{3}} = \frac{4}{16\sqrt{3}} = \frac{1}{4\sqrt{3}}$$ So the answer is D. \(\frac{1}{4\sqrt{3}}\). |
B |
| doubao | First, simplify the radicals: $\sqrt{108}=\sqrt{36\times3}=6\sqrt{3}$, $\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}$, $\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}$. So the denominator becomes $6\sqrt{3}+2\times2\sqrt{3}+2\times3\sqrt{3}=6\sqrt{3}+4\sqrt{3}+6\sqrt{3}=16\sqrt{3}$. Then the fraction becomes $\frac{4}{16\sqrt{3}}=\frac{1}{4\sqrt{3}}$. So the answer is D. |
B |
| minimax | To simplify the fraction by rationalizing the denominator, we first need to simplify the denominator itself. Let's start by simplifying each square root term: $$\sqrt{108} = \sqrt{36 \cdot 3} = \sqrt{36} \cdot \sqrt{3} = 6\sqrt{3}$$ $$\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}$$ $$\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}$$ Now, let's substitute these simplified forms back into the original expression: $$\frac{4}{\sqrt{108}+2\sqrt{12}+2\sqrt{27}} = \frac{4}{6\sqrt{3} + 2(2\sqrt{3}) + 2(3\sqrt{3})}$$ $$= \frac{4}{6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3}}$$ $$= \frac{4}{16\sqrt{3}}$$ Now, we can simplify the fraction by dividing both the numerator and the denominator by 4: $$= \frac{4/4}{16\sqrt{3}/4}$$ $$= \frac{1}{4\sqrt{3}}$$ Therefore, the correct answer is: D. \frac{1}{4\sqrt{3}} |
B |
| kimi | To simplify the fraction by rationalizing the denominator, we first need to simplify the square roots in the denominator and then rationalize it. The denominator is $\sqrt{108} + 2\sqrt{12} + 2\sqrt{27}$. Let's simplify each square root: - $\sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3}$ - $2\sqrt{12} = 2\sqrt{4 \cdot 3} = 2 \cdot 2\sqrt{3} = 4\sqrt{3}$ - $2\sqrt{27} = 2\sqrt{9 \cdot 3} = 2 \cdot 3\sqrt{3} = 6\sqrt{3}$ Now, let's add them up: $6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3} = 16\sqrt{3}$ So the fraction becomes: $$\frac{4}{16\sqrt{3}}$$ Now, we can simplify this fraction by dividing both the numerator and the denominator by 4: $$\frac{4}{16\sqrt{3}} = \frac{1}{4\sqrt{3}}$$ To rationalize the denominator, we multiply both the numerator and the denominator by $\sqrt{3}$: $$\frac{1}{4\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{12}$$ So the simplified and rationalized fraction is $\frac{\sqrt{3}}{12}$. Answer: B. $\frac{\sqrt{3}}{12}$ |
B |
| wenxinyiyan | To simplify the given fraction, we first need to rationalize the denominator. We can do this by multiplying the numerator and denominator by the conjugate of the denominator. However, in this case, the denominator is a sum of square roots, so we need to find a way to simplify it. First, let's simplify the square roots in the denominator: √108 = √(36 * 3) = 6√3 2√12 = 2√(4 * 3) = 4√3 2√27 = 2√(9 * 3) = 6√3 Now, the denominator becomes: 6√3 + 4√3 + 6√3 = 16√3 The fraction is now: 4 / (16√3) To rationalize the denominator, we multiply the numerator and denominator by √3: (4 * √3) / (16√3 * √3) = (4√3) / 48 Simplifying the fraction gives: √3 / 12 So, the correct answer is B. |
B |
| zhipu | To simplify the fraction by rationalizing the denominator, we start with: $$\frac{4}{\sqrt{108} + 2\sqrt{12} + 2\sqrt{27}}.$$ First, we simplify each term in the denominator: - \(\sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3}\) - \(2\sqrt{12} = 2\sqrt{4 \cdot 3} = 2 \cdot 2\sqrt{3} = 4\sqrt{3}\) - \(2\sqrt{27} = 2\sqrt{9 \cdot 3} = 2 \cdot 3\sqrt{3} = 6\sqrt{3}\) Substituting these simplified terms back into the denominator, we get: $$\frac{4}{6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3}}.$$ Now, combine the like terms in the denominator: $$6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3} = (6 + 4 + 6)\sqrt{3} = 16\sqrt{3}.$$ So the fraction becomes: $$\frac{4}{16\sqrt{3}}.$$ We can simplify this fraction by dividing the numerator and the denominator by 4: $$\frac{4 \div 4}{16\sqrt{3} \div 4} = \frac{1}{4\sqrt{3}}.$$ To rationalize the denominator, multiply the numerator and the denominator by \(\sqrt{3}\): $$\frac{1 \cdot \sqrt{3}}{4\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{3}}{4 \cdot 3} = \frac{\sqrt{3}}{12}.$$ Thus, the simplified form of the given fraction is: $$\frac{\sqrt{3}}{12}.$$ Therefore, the correct answer is: B. \(\frac{\sqrt{3}}{12}\). |
B |
| gpto1 | To solve this problem, we'll simplify each radical and then rationalize the denominator. 1. **Simplify the radicals:** - \(\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\) - \(2\sqrt{12} = 2 \times \sqrt{4 \times 3} = 2 \times 2\sqrt{3} = 4\sqrt{3}\) - \(2\sqrt{27} = 2 \times \sqrt{9 \times 3} = 2 \times 3\sqrt{3} = 6\sqrt{3}\) 2. **Combine like terms in the denominator:** \(6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3} = 16\sqrt{3}\) 3. **Simplify the fraction:** \(\frac{4}{16\sqrt{3}} = \frac{1}{4\sqrt{3}}\) 4. **Rationalize the denominator:** \(\frac{1}{4\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{12}\) So, the simplified fraction is \(\frac{\sqrt{3}}{12}\), which corresponds to option **B**. **Answer: B** |
B |
| deepseekv3 | To simplify the given expression, follow these steps: **Given Expression:** \[ \frac{4}{\sqrt{108} + 2\sqrt{12} + 2\sqrt{27}} \] **Step 1: Simplify each square root in the denominator:** \[ \sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3} \] \[ \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \] \[ \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \] **Step 2: Substitute the simplified square roots back into the denominator:** \[ \sqrt{108} + 2\sqrt{12} + 2\sqrt{27} = 6\sqrt{3} + 2(2\sqrt{3}) + 2(3\sqrt{3}) = 6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3} \] **Step 3: Combine like terms in the denominator:** \[ 6\sqrt{3} + 4\sqrt{3} + 6\sqrt{3} = 16\sqrt{3} \] **Step 4: Rewrite the original expression with the simplified denominator:** \[ \frac{4}{16\sqrt{3}} = \frac{1}{4\sqrt{3}} \] **Step 5: Rationalize the denominator:** \[ \frac{1}{4\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{12} \] **Final Answer:** \[ \boxed{B} \] |
B |