Answer
$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}-\sqrt{5}}=3\sqrt{2}+\sqrt{15}+2\sqrt{3}+\sqrt{10}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}-\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}-\sqrt{5}}\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3\sqrt{2}+\sqrt{15}+2\sqrt{3}+\sqrt{10}}{6+\sqrt{30}-\sqrt{30}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3\sqrt{2}+\sqrt{15}+2\sqrt{3}+\sqrt{10}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}3\sqrt{2}+\sqrt{15}+2\sqrt{3}+\sqrt{10}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{6} + \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3} + \sqrt{2}\right) } \cdot \left( \sqrt{6} + \sqrt{5}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{6}+\color{blue}{ \sqrt{3}} \cdot \sqrt{5}+\color{blue}{ \sqrt{2}} \cdot \sqrt{6}+\color{blue}{ \sqrt{2}} \cdot \sqrt{5} = \\ = 3 \sqrt{2} + \sqrt{15} + 2 \sqrt{3} + \sqrt{10} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{6}- \sqrt{5}\right) } \cdot \left( \sqrt{6} + \sqrt{5}\right) = \color{blue}{ \sqrt{6}} \cdot \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot \sqrt{5}\color{blue}{- \sqrt{5}} \cdot \sqrt{6}\color{blue}{- \sqrt{5}} \cdot \sqrt{5} = \\ = 6 + \sqrt{30}- \sqrt{30}-5 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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