Answer
$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{12}-\sqrt{2}}=\frac{8+3\sqrt{6}}{10}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}+\sqrt{2}}{\sqrt{12}-\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}+\sqrt{2}}{\sqrt{12}-\sqrt{2}}\frac{\sqrt{12}+\sqrt{2}}{\sqrt{12}+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6+\sqrt{6}+2\sqrt{6}+2}{12+2\sqrt{6}-2\sqrt{6}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8+3\sqrt{6}}{10}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{12} + \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3} + \sqrt{2}\right) } \cdot \left( \sqrt{12} + \sqrt{2}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{12}+\color{blue}{ \sqrt{3}} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot \sqrt{12}+\color{blue}{ \sqrt{2}} \cdot \sqrt{2} = \\ = 6 + \sqrt{6} + 2 \sqrt{6} + 2 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{12}- \sqrt{2}\right) } \cdot \left( \sqrt{12} + \sqrt{2}\right) = \color{blue}{ \sqrt{12}} \cdot \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot \sqrt{12}\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 12 + 2 \sqrt{6}- 2 \sqrt{6}-2 $$ |
| ③ | Simplify numerator and denominator |
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