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