Answer
$$\frac{\sqrt{12}+\sqrt{8}}{\sqrt{8}+\sqrt{3}}=\frac{2\sqrt{6}+2}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{12}+\sqrt{8}}{\sqrt{8}+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{12}+\sqrt{8}}{\sqrt{8}+\sqrt{3}}\frac{\sqrt{8}-\sqrt{3}}{\sqrt{8}-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4\sqrt{6}-6+8-2\sqrt{6}}{8-2\sqrt{6}+2\sqrt{6}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{6}+2}{5}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{8}- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{12} + \sqrt{8}\right) } \cdot \left( \sqrt{8}- \sqrt{3}\right) = \color{blue}{ \sqrt{12}} \cdot \sqrt{8}+\color{blue}{ \sqrt{12}} \cdot- \sqrt{3}+\color{blue}{ \sqrt{8}} \cdot \sqrt{8}+\color{blue}{ \sqrt{8}} \cdot- \sqrt{3} = \\ = 4 \sqrt{6}-6 + 8- 2 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{8} + \sqrt{3}\right) } \cdot \left( \sqrt{8}- \sqrt{3}\right) = \color{blue}{ \sqrt{8}} \cdot \sqrt{8}+\color{blue}{ \sqrt{8}} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot \sqrt{8}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 8- 2 \sqrt{6} + 2 \sqrt{6}-3 $$ |
| ③ | Simplify numerator and denominator |
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