Answer
$$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\frac{\sqrt{8}-\sqrt{12}}{\sqrt{8}-\sqrt{12}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{16-8\sqrt{6}+8\sqrt{6}-24}{8-4\sqrt{6}+4\sqrt{6}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-8}{-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{8}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}} \frac{ 8 : \color{orangered}{ 4 } }{ 4 : \color{orangered}{ 4 }} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{2}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}2\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{8}- \sqrt{12}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{32} + \sqrt{48}\right) } \cdot \left( \sqrt{8}- \sqrt{12}\right) = \color{blue}{ \sqrt{32}} \cdot \sqrt{8}+\color{blue}{ \sqrt{32}} \cdot- \sqrt{12}+\color{blue}{ \sqrt{48}} \cdot \sqrt{8}+\color{blue}{ \sqrt{48}} \cdot- \sqrt{12} = \\ = 16- 8 \sqrt{6} + 8 \sqrt{6}-24 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{8} + \sqrt{12}\right) } \cdot \left( \sqrt{8}- \sqrt{12}\right) = \color{blue}{ \sqrt{8}} \cdot \sqrt{8}+\color{blue}{ \sqrt{8}} \cdot- \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot \sqrt{8}+\color{blue}{ \sqrt{12}} \cdot- \sqrt{12} = \\ = 8- 4 \sqrt{6} + 4 \sqrt{6}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | cancel two minus signs |
| ⑤ | Divide both the top and bottom numbers by $ \color{orangered}{ 4 } $. |
| ⑥ | Remove 1 from denominator. |
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