Answer
$$\frac{2-\sqrt{5}}{\sqrt{5}-4}=\frac{2\sqrt{5}-3}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2-\sqrt{5}}{\sqrt{5}-4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2-\sqrt{5}}{\sqrt{5}-4}\frac{\sqrt{5}+4}{\sqrt{5}+4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{5}+8-5-4\sqrt{5}}{5+4\sqrt{5}-4\sqrt{5}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-2\sqrt{5}+3}{-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2\sqrt{5}-3}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{5} + 4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2- \sqrt{5}\right) } \cdot \left( \sqrt{5} + 4\right) = \color{blue}{2} \cdot \sqrt{5}+\color{blue}{2} \cdot4\color{blue}{- \sqrt{5}} \cdot \sqrt{5}\color{blue}{- \sqrt{5}} \cdot4 = \\ = 2 \sqrt{5} + 8-5- 4 \sqrt{5} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{5}-4\right) } \cdot \left( \sqrt{5} + 4\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot4\color{blue}{-4} \cdot \sqrt{5}\color{blue}{-4} \cdot4 = \\ = 5 + 4 \sqrt{5}- 4 \sqrt{5}-16 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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