Answer
$$\frac{2-\sqrt{3}}{1+\sqrt{2}}=-2+2\sqrt{2}+\sqrt{3}-\sqrt{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2-\sqrt{3}}{1+\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2-\sqrt{3}}{1+\sqrt{2}}\frac{1-\sqrt{2}}{1-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2-2\sqrt{2}-\sqrt{3}+\sqrt{6}}{1-\sqrt{2}+\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2-2\sqrt{2}-\sqrt{3}+\sqrt{6}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2+2\sqrt{2}+\sqrt{3}-\sqrt{6}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-2+2\sqrt{2}+\sqrt{3}-\sqrt{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 1- \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2- \sqrt{3}\right) } \cdot \left( 1- \sqrt{2}\right) = \color{blue}{2} \cdot1+\color{blue}{2} \cdot- \sqrt{2}\color{blue}{- \sqrt{3}} \cdot1\color{blue}{- \sqrt{3}} \cdot- \sqrt{2} = \\ = 2- 2 \sqrt{2}- \sqrt{3} + \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 1 + \sqrt{2}\right) } \cdot \left( 1- \sqrt{2}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot1+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 1- \sqrt{2} + \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
| ⑤ | Remove 1 from denominator. |
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