Question
Answer
$$\dfrac{(3-2\sqrt{2})^2(3-2\sqrt{2})^2+1}{(3-2\sqrt{2})^2}=34$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(3-2\sqrt{2})^2(3-2\sqrt{2})^2+1}{(3-2\sqrt{2})^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{(9-6\sqrt{2}-6\sqrt{2}+8)(9-6\sqrt{2}-6\sqrt{2}+8)+1}{(3-2\sqrt{2})^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{(17-12\sqrt{2})\cdot(17-12\sqrt{2})+1}{(3-2\sqrt{2})^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{289-204\sqrt{2}-204\sqrt{2}+288+1}{(3-2\sqrt{2})^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{577-408\sqrt{2}+1}{(3-2\sqrt{2})^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{578-408\sqrt{2}}{9-6\sqrt{2}-6\sqrt{2}+8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}\frac{578-408\sqrt{2}}{17-12\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}\frac{578-408\sqrt{2}}{17-12\sqrt{2}}\frac{17+12\sqrt{2}}{17+12\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle11}{\textcircled {11}} } }}}\frac{9826+6936\sqrt{2}-6936\sqrt{2}-9792}{289+204\sqrt{2}-204\sqrt{2}-288} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle12}{\textcircled {12}} } }}}\frac{34}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle13}{\textcircled {13}} } }}}34\end{aligned} $$ | |
| ① | $$ (3-2\sqrt{2})^2 = \left( 3- 2 \sqrt{2} \right) \cdot \left( 3- 2 \sqrt{2} \right) = 9- 6 \sqrt{2}- 6 \sqrt{2} + 8 $$ |
| ② | $$ (3-2\sqrt{2})^2 = \left( 3- 2 \sqrt{2} \right) \cdot \left( 3- 2 \sqrt{2} \right) = 9- 6 \sqrt{2}- 6 \sqrt{2} + 8 $$ |
| ③ | Combine like terms |
| ④ | Combine like terms |
| ⑤ | $$ \color{blue}{ \left( 17- 12 \sqrt{2}\right) } \cdot \left( 17- 12 \sqrt{2}\right) = \color{blue}{17} \cdot17+\color{blue}{17} \cdot- 12 \sqrt{2}\color{blue}{- 12 \sqrt{2}} \cdot17\color{blue}{- 12 \sqrt{2}} \cdot- 12 \sqrt{2} = \\ = 289- 204 \sqrt{2}- 204 \sqrt{2} + 288 $$ |
| ⑥ | Combine like terms |
| ⑦ | Combine like terms |
| ⑧ | $$ (3-2\sqrt{2})^2 = \left( 3- 2 \sqrt{2} \right) \cdot \left( 3- 2 \sqrt{2} \right) = 9- 6 \sqrt{2}- 6 \sqrt{2} + 8 $$ |
| ⑨ | Simplify numerator and denominator |
| ⑩ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 17 + 12 \sqrt{2}} $$. |
| ⑪ | Multiply in a numerator. $$ \color{blue}{ \left( 578- 408 \sqrt{2}\right) } \cdot \left( 17 + 12 \sqrt{2}\right) = \color{blue}{578} \cdot17+\color{blue}{578} \cdot 12 \sqrt{2}\color{blue}{- 408 \sqrt{2}} \cdot17\color{blue}{- 408 \sqrt{2}} \cdot 12 \sqrt{2} = \\ = 9826 + 6936 \sqrt{2}- 6936 \sqrt{2}-9792 $$ Simplify denominator. $$ \color{blue}{ \left( 17- 12 \sqrt{2}\right) } \cdot \left( 17 + 12 \sqrt{2}\right) = \color{blue}{17} \cdot17+\color{blue}{17} \cdot 12 \sqrt{2}\color{blue}{- 12 \sqrt{2}} \cdot17\color{blue}{- 12 \sqrt{2}} \cdot 12 \sqrt{2} = \\ = 289 + 204 \sqrt{2}- 204 \sqrt{2}-288 $$ |
| ⑫ | Simplify numerator and denominator |
| ⑬ | Remove 1 from denominator. |
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