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