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