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