Answer
$$\frac{\sqrt{3}+1}{2\sqrt{2}-\sqrt{3}}=\frac{2\sqrt{6}+3+2\sqrt{2}+\sqrt{3}}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}+1}{2\sqrt{2}-\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}+1}{2\sqrt{2}-\sqrt{3}}\frac{2\sqrt{2}+\sqrt{3}}{2\sqrt{2}+\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{6}+3+2\sqrt{2}+\sqrt{3}}{8+2\sqrt{6}-2\sqrt{6}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{6}+3+2\sqrt{2}+\sqrt{3}}{5}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{2} + \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3} + 1\right) } \cdot \left( 2 \sqrt{2} + \sqrt{3}\right) = \color{blue}{ \sqrt{3}} \cdot 2 \sqrt{2}+\color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{1} \cdot 2 \sqrt{2}+\color{blue}{1} \cdot \sqrt{3} = \\ = 2 \sqrt{6} + 3 + 2 \sqrt{2} + \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{2}- \sqrt{3}\right) } \cdot \left( 2 \sqrt{2} + \sqrt{3}\right) = \color{blue}{ 2 \sqrt{2}} \cdot 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot \sqrt{3}\color{blue}{- \sqrt{3}} \cdot 2 \sqrt{2}\color{blue}{- \sqrt{3}} \cdot \sqrt{3} = \\ = 8 + 2 \sqrt{6}- 2 \sqrt{6}-3 $$ |
| ③ | Simplify numerator and denominator |
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