Answer
$$\frac{1}{\sqrt{5}-2}=\sqrt{5}+2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{5}-2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{\sqrt{5}-2}\frac{\sqrt{5}+2}{\sqrt{5}+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{5}+2}{5+2\sqrt{5}-2\sqrt{5}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{5}+2}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\sqrt{5}+2\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{5} + 2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( \sqrt{5} + 2\right) = \color{blue}{1} \cdot \sqrt{5}+\color{blue}{1} \cdot2 = \\ = \sqrt{5} + 2 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{5}-2\right) } \cdot \left( \sqrt{5} + 2\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot2\color{blue}{-2} \cdot \sqrt{5}\color{blue}{-2} \cdot2 = \\ = 5 + 2 \sqrt{5}- 2 \sqrt{5}-4 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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