Answer
$$\frac{\sqrt{2}+\sqrt{3}}{\sqrt{18}-\sqrt{12}}=\frac{12+5\sqrt{6}}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{2}+\sqrt{3}}{\sqrt{18}-\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{2}+\sqrt{3}}{\sqrt{18}-\sqrt{12}}\frac{\sqrt{18}+\sqrt{12}}{\sqrt{18}+\sqrt{12}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6+2\sqrt{6}+3\sqrt{6}+6}{18+6\sqrt{6}-6\sqrt{6}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12+5\sqrt{6}}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{18} + \sqrt{12}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{2} + \sqrt{3}\right) } \cdot \left( \sqrt{18} + \sqrt{12}\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{18}+\color{blue}{ \sqrt{2}} \cdot \sqrt{12}+\color{blue}{ \sqrt{3}} \cdot \sqrt{18}+\color{blue}{ \sqrt{3}} \cdot \sqrt{12} = \\ = 6 + 2 \sqrt{6} + 3 \sqrt{6} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{18}- \sqrt{12}\right) } \cdot \left( \sqrt{18} + \sqrt{12}\right) = \color{blue}{ \sqrt{18}} \cdot \sqrt{18}+\color{blue}{ \sqrt{18}} \cdot \sqrt{12}\color{blue}{- \sqrt{12}} \cdot \sqrt{18}\color{blue}{- \sqrt{12}} \cdot \sqrt{12} = \\ = 18 + 6 \sqrt{6}- 6 \sqrt{6}-12 $$ |
| ③ | Simplify numerator and denominator |
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