Answer
$$\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{114-41\sqrt{6}}{30}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}\frac{\sqrt{48}-\sqrt{18}}{\sqrt{48}-\sqrt{18}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{84-21\sqrt{6}-20\sqrt{6}+30}{48-12\sqrt{6}+12\sqrt{6}-18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{114-41\sqrt{6}}{30}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{48}- \sqrt{18}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 7 \sqrt{3}- 5 \sqrt{2}\right) } \cdot \left( \sqrt{48}- \sqrt{18}\right) = \color{blue}{ 7 \sqrt{3}} \cdot \sqrt{48}+\color{blue}{ 7 \sqrt{3}} \cdot- \sqrt{18}\color{blue}{- 5 \sqrt{2}} \cdot \sqrt{48}\color{blue}{- 5 \sqrt{2}} \cdot- \sqrt{18} = \\ = 84- 21 \sqrt{6}- 20 \sqrt{6} + 30 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{48} + \sqrt{18}\right) } \cdot \left( \sqrt{48}- \sqrt{18}\right) = \color{blue}{ \sqrt{48}} \cdot \sqrt{48}+\color{blue}{ \sqrt{48}} \cdot- \sqrt{18}+\color{blue}{ \sqrt{18}} \cdot \sqrt{48}+\color{blue}{ \sqrt{18}} \cdot- \sqrt{18} = \\ = 48- 12 \sqrt{6} + 12 \sqrt{6}-18 $$ |
| ③ | Simplify numerator and denominator |
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