Answer
$$\frac{\sqrt{6}+9}{\sqrt{6}-1}=3+2\sqrt{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{6}+9}{\sqrt{6}-1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{6}+9}{\sqrt{6}-1}\frac{\sqrt{6}+1}{\sqrt{6}+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6+\sqrt{6}+9\sqrt{6}+9}{6+\sqrt{6}-\sqrt{6}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{15+10\sqrt{6}}{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{3+2\sqrt{6}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}3+2\sqrt{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{6} + 1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{6} + 9\right) } \cdot \left( \sqrt{6} + 1\right) = \color{blue}{ \sqrt{6}} \cdot \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot1+\color{blue}{9} \cdot \sqrt{6}+\color{blue}{9} \cdot1 = \\ = 6 + \sqrt{6} + 9 \sqrt{6} + 9 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{6}-1\right) } \cdot \left( \sqrt{6} + 1\right) = \color{blue}{ \sqrt{6}} \cdot \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot1\color{blue}{-1} \cdot \sqrt{6}\color{blue}{-1} \cdot1 = \\ = 6 + \sqrt{6}- \sqrt{6}-1 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 5. |
| ⑤ | Remove 1 from denominator. |
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