Answer
$$\frac{1}{(4-\sqrt{7})^2}=\frac{23+8\sqrt{7}}{81}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{(4-\sqrt{7})^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{16-4\sqrt{7}-4\sqrt{7}+7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1}{23-8\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1}{23-8\sqrt{7}}\frac{23+8\sqrt{7}}{23+8\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{23+8\sqrt{7}}{529+184\sqrt{7}-184\sqrt{7}-448} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{23+8\sqrt{7}}{81}\end{aligned} $$ | |
| ① | $$ (4-\sqrt{7})^2 = \left( 4- \sqrt{7} \right) \cdot \left( 4- \sqrt{7} \right) = 16- 4 \sqrt{7}- 4 \sqrt{7} + 7 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 23 + 8 \sqrt{7}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 23 + 8 \sqrt{7}\right) = \color{blue}{1} \cdot23+\color{blue}{1} \cdot 8 \sqrt{7} = \\ = 23 + 8 \sqrt{7} $$ Simplify denominator. $$ \color{blue}{ \left( 23- 8 \sqrt{7}\right) } \cdot \left( 23 + 8 \sqrt{7}\right) = \color{blue}{23} \cdot23+\color{blue}{23} \cdot 8 \sqrt{7}\color{blue}{- 8 \sqrt{7}} \cdot23\color{blue}{- 8 \sqrt{7}} \cdot 8 \sqrt{7} = \\ = 529 + 184 \sqrt{7}- 184 \sqrt{7}-448 $$ |
| ⑤ | Simplify numerator and denominator |
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