Answer
$$\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}=\frac{60+24\sqrt{6}}{9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}\frac{\sqrt{27}+\sqrt{18}}{\sqrt{27}+\sqrt{18}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{36+12\sqrt{6}+12\sqrt{6}+24}{27+9\sqrt{6}-9\sqrt{6}-18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{60+24\sqrt{6}}{9}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{27} + \sqrt{18}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{48} + \sqrt{32}\right) } \cdot \left( \sqrt{27} + \sqrt{18}\right) = \color{blue}{ \sqrt{48}} \cdot \sqrt{27}+\color{blue}{ \sqrt{48}} \cdot \sqrt{18}+\color{blue}{ \sqrt{32}} \cdot \sqrt{27}+\color{blue}{ \sqrt{32}} \cdot \sqrt{18} = \\ = 36 + 12 \sqrt{6} + 12 \sqrt{6} + 24 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{27}- \sqrt{18}\right) } \cdot \left( \sqrt{27} + \sqrt{18}\right) = \color{blue}{ \sqrt{27}} \cdot \sqrt{27}+\color{blue}{ \sqrt{27}} \cdot \sqrt{18}\color{blue}{- \sqrt{18}} \cdot \sqrt{27}\color{blue}{- \sqrt{18}} \cdot \sqrt{18} = \\ = 27 + 9 \sqrt{6}- 9 \sqrt{6}-18 $$ |
| ③ | Simplify numerator and denominator |
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