Answer
$$\frac{3\sqrt{5}}{9-\sqrt{5}}=\frac{27\sqrt{5}+15}{76}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{5}}{9-\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3\sqrt{5}}{9-\sqrt{5}}\frac{9+\sqrt{5}}{9+\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{27\sqrt{5}+15}{81+9\sqrt{5}-9\sqrt{5}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{27\sqrt{5}+15}{76}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 9 + \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 \sqrt{5} } \cdot \left( 9 + \sqrt{5}\right) = \color{blue}{ 3 \sqrt{5}} \cdot9+\color{blue}{ 3 \sqrt{5}} \cdot \sqrt{5} = \\ = 27 \sqrt{5} + 15 $$ Simplify denominator. $$ \color{blue}{ \left( 9- \sqrt{5}\right) } \cdot \left( 9 + \sqrt{5}\right) = \color{blue}{9} \cdot9+\color{blue}{9} \cdot \sqrt{5}\color{blue}{- \sqrt{5}} \cdot9\color{blue}{- \sqrt{5}} \cdot \sqrt{5} = \\ = 81 + 9 \sqrt{5}- 9 \sqrt{5}-5 $$ |
| ③ | Simplify numerator and denominator |
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