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