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