Answer
$$\frac{9-\sqrt{6}}{8+3\sqrt{6}}=\frac{18-7\sqrt{6}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{9-\sqrt{6}}{8+3\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9-\sqrt{6}}{8+3\sqrt{6}}\frac{8-3\sqrt{6}}{8-3\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{72-27\sqrt{6}-8\sqrt{6}+18}{64-24\sqrt{6}+24\sqrt{6}-54} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{90-35\sqrt{6}}{10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{18-7\sqrt{6}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 8- 3 \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 9- \sqrt{6}\right) } \cdot \left( 8- 3 \sqrt{6}\right) = \color{blue}{9} \cdot8+\color{blue}{9} \cdot- 3 \sqrt{6}\color{blue}{- \sqrt{6}} \cdot8\color{blue}{- \sqrt{6}} \cdot- 3 \sqrt{6} = \\ = 72- 27 \sqrt{6}- 8 \sqrt{6} + 18 $$ Simplify denominator. $$ \color{blue}{ \left( 8 + 3 \sqrt{6}\right) } \cdot \left( 8- 3 \sqrt{6}\right) = \color{blue}{8} \cdot8+\color{blue}{8} \cdot- 3 \sqrt{6}+\color{blue}{ 3 \sqrt{6}} \cdot8+\color{blue}{ 3 \sqrt{6}} \cdot- 3 \sqrt{6} = \\ = 64- 24 \sqrt{6} + 24 \sqrt{6}-54 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 5. |
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