Answer
$$\frac{1}{\sqrt{8}+\sqrt{6}}=\frac{2\sqrt{2}-\sqrt{6}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{8}+\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{\sqrt{8}+\sqrt{6}}\frac{\sqrt{8}-\sqrt{6}}{\sqrt{8}-\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{2}-\sqrt{6}}{8-4\sqrt{3}+4\sqrt{3}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{2}-\sqrt{6}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{8}- \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( \sqrt{8}- \sqrt{6}\right) = \color{blue}{1} \cdot \sqrt{8}+\color{blue}{1} \cdot- \sqrt{6} = \\ = 2 \sqrt{2}- \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{8} + \sqrt{6}\right) } \cdot \left( \sqrt{8}- \sqrt{6}\right) = \color{blue}{ \sqrt{8}} \cdot \sqrt{8}+\color{blue}{ \sqrt{8}} \cdot- \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot \sqrt{8}+\color{blue}{ \sqrt{6}} \cdot- \sqrt{6} = \\ = 8- 4 \sqrt{3} + 4 \sqrt{3}-6 $$ |
| ③ | Simplify numerator and denominator |
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