Answer
$$\frac{\sqrt{12}}{\sqrt{3}+2}=-6+4\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{12}}{\sqrt{3}+2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{12}}{\sqrt{3}+2}\frac{\sqrt{3}-2}{\sqrt{3}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6-4\sqrt{3}}{3-2\sqrt{3}+2\sqrt{3}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6-4\sqrt{3}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-6+4\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-6+4\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}-2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{12} } \cdot \left( \sqrt{3}-2\right) = \color{blue}{ \sqrt{12}} \cdot \sqrt{3}+\color{blue}{ \sqrt{12}} \cdot-2 = \\ = 6- 4 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3} + 2\right) } \cdot \left( \sqrt{3}-2\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot-2+\color{blue}{2} \cdot \sqrt{3}+\color{blue}{2} \cdot-2 = \\ = 3- 2 \sqrt{3} + 2 \sqrt{3}-4 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
| ⑤ | Remove 1 from denominator. |
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