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