Answer
$$\frac{1}{(5-\sqrt{2})^2}=\frac{27+10\sqrt{2}}{529}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{(5-\sqrt{2})^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{25-5\sqrt{2}-5\sqrt{2}+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1}{27-10\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1}{27-10\sqrt{2}}\frac{27+10\sqrt{2}}{27+10\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{27+10\sqrt{2}}{729+270\sqrt{2}-270\sqrt{2}-200} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{27+10\sqrt{2}}{529}\end{aligned} $$ | |
| ① | $$ (5-\sqrt{2})^2 = \left( 5- \sqrt{2} \right) \cdot \left( 5- \sqrt{2} \right) = 25- 5 \sqrt{2}- 5 \sqrt{2} + 2 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 27 + 10 \sqrt{2}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 27 + 10 \sqrt{2}\right) = \color{blue}{1} \cdot27+\color{blue}{1} \cdot 10 \sqrt{2} = \\ = 27 + 10 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 27- 10 \sqrt{2}\right) } \cdot \left( 27 + 10 \sqrt{2}\right) = \color{blue}{27} \cdot27+\color{blue}{27} \cdot 10 \sqrt{2}\color{blue}{- 10 \sqrt{2}} \cdot27\color{blue}{- 10 \sqrt{2}} \cdot 10 \sqrt{2} = \\ = 729 + 270 \sqrt{2}- 270 \sqrt{2}-200 $$ |
| ⑤ | Simplify numerator and denominator |
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