Answer
$$\frac{1}{(2+\sqrt{5})^2}=9-4\sqrt{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{(2+\sqrt{5})^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{4+2\sqrt{5}+2\sqrt{5}+5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1}{9+4\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1}{9+4\sqrt{5}}\frac{9-4\sqrt{5}}{9-4\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{9-4\sqrt{5}}{81-36\sqrt{5}+36\sqrt{5}-80} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{9-4\sqrt{5}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}9-4\sqrt{5}\end{aligned} $$ | |
| ① | $$ (2+\sqrt{5})^2 = \left( 2 + \sqrt{5} \right) \cdot \left( 2 + \sqrt{5} \right) = 4 + 2 \sqrt{5} + 2 \sqrt{5} + 5 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 9- 4 \sqrt{5}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 9- 4 \sqrt{5}\right) = \color{blue}{1} \cdot9+\color{blue}{1} \cdot- 4 \sqrt{5} = \\ = 9- 4 \sqrt{5} $$ Simplify denominator. $$ \color{blue}{ \left( 9 + 4 \sqrt{5}\right) } \cdot \left( 9- 4 \sqrt{5}\right) = \color{blue}{9} \cdot9+\color{blue}{9} \cdot- 4 \sqrt{5}+\color{blue}{ 4 \sqrt{5}} \cdot9+\color{blue}{ 4 \sqrt{5}} \cdot- 4 \sqrt{5} = \\ = 81- 36 \sqrt{5} + 36 \sqrt{5}-80 $$ |
| ⑤ | Simplify numerator and denominator |
| ⑥ | Remove 1 from denominator. |
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