Answer
$$\frac{\sqrt{3}+6}{\sqrt{2}-4}=-\frac{\sqrt{6}+4\sqrt{3}+6\sqrt{2}+24}{14}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}+6}{\sqrt{2}-4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}+6}{\sqrt{2}-4}\frac{\sqrt{2}+4}{\sqrt{2}+4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{6}+4\sqrt{3}+6\sqrt{2}+24}{2+4\sqrt{2}-4\sqrt{2}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{6}+4\sqrt{3}+6\sqrt{2}+24}{-14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{\sqrt{6}+4\sqrt{3}+6\sqrt{2}+24}{14}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2} + 4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3} + 6\right) } \cdot \left( \sqrt{2} + 4\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{2}+\color{blue}{ \sqrt{3}} \cdot4+\color{blue}{6} \cdot \sqrt{2}+\color{blue}{6} \cdot4 = \\ = \sqrt{6} + 4 \sqrt{3} + 6 \sqrt{2} + 24 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{2}-4\right) } \cdot \left( \sqrt{2} + 4\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot4\color{blue}{-4} \cdot \sqrt{2}\color{blue}{-4} \cdot4 = \\ = 2 + 4 \sqrt{2}- 4 \sqrt{2}-16 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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