Answer
$$\frac{9-\sqrt{6}}{3+\sqrt{6}}=11-4\sqrt{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{9-\sqrt{6}}{3+\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9-\sqrt{6}}{3+\sqrt{6}}\frac{3-\sqrt{6}}{3-\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{27-9\sqrt{6}-3\sqrt{6}+6}{9-3\sqrt{6}+3\sqrt{6}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{33-12\sqrt{6}}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{11-4\sqrt{6}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}11-4\sqrt{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 9- \sqrt{6}\right) } \cdot \left( 3- \sqrt{6}\right) = \color{blue}{9} \cdot3+\color{blue}{9} \cdot- \sqrt{6}\color{blue}{- \sqrt{6}} \cdot3\color{blue}{- \sqrt{6}} \cdot- \sqrt{6} = \\ = 27- 9 \sqrt{6}- 3 \sqrt{6} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( 3 + \sqrt{6}\right) } \cdot \left( 3- \sqrt{6}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot3+\color{blue}{ \sqrt{6}} \cdot- \sqrt{6} = \\ = 9- 3 \sqrt{6} + 3 \sqrt{6}-6 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
| ⑤ | Remove 1 from denominator. |
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