Question
Answer
$$\dfrac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2}=17+12\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2+\sqrt{2}+\sqrt{2}+1}{2-\sqrt{2}-\sqrt{2}+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3+2\sqrt{2}}{3-2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\frac{3+2\sqrt{2}}{3+2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{9+6\sqrt{2}+6\sqrt{2}+8}{9+6\sqrt{2}-6\sqrt{2}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{17+12\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}17+12\sqrt{2}\end{aligned} $$ | |
| ① | $$ (\sqrt{2}+1)^2 = \left( \sqrt{2} + 1 \right) \cdot \left( \sqrt{2} + 1 \right) = 2 + \sqrt{2} + \sqrt{2} + 1 $$ |
| ② | $$ (\sqrt{2}-1)^2 = \left( \sqrt{2}-1 \right) \cdot \left( \sqrt{2}-1 \right) = 2- \sqrt{2}- \sqrt{2} + 1 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 + 2 \sqrt{2}} $$. |
| ⑤ | Multiply in a numerator. $$ \color{blue}{ \left( 3 + 2 \sqrt{2}\right) } \cdot \left( 3 + 2 \sqrt{2}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot3+\color{blue}{ 2 \sqrt{2}} \cdot 2 \sqrt{2} = \\ = 9 + 6 \sqrt{2} + 6 \sqrt{2} + 8 $$ Simplify denominator. $$ \color{blue}{ \left( 3- 2 \sqrt{2}\right) } \cdot \left( 3 + 2 \sqrt{2}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot 2 \sqrt{2}\color{blue}{- 2 \sqrt{2}} \cdot3\color{blue}{- 2 \sqrt{2}} \cdot 2 \sqrt{2} = \\ = 9 + 6 \sqrt{2}- 6 \sqrt{2}-8 $$ |
| ⑥ | Simplify numerator and denominator |
| ⑦ | Remove 1 from denominator. |
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