Answer
$$\frac{2}{2+2\sqrt{2}}=-1+\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{2+2\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{2+2\sqrt{2}}\frac{2-2\sqrt{2}}{2-2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4-4\sqrt{2}}{4-4\sqrt{2}+4\sqrt{2}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4-4\sqrt{2}}{-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{1-\sqrt{2}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-1+\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-1+\sqrt{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- 2 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( 2- 2 \sqrt{2}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- 2 \sqrt{2} = \\ = 4- 4 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 2 + 2 \sqrt{2}\right) } \cdot \left( 2- 2 \sqrt{2}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot2+\color{blue}{ 2 \sqrt{2}} \cdot- 2 \sqrt{2} = \\ = 4- 4 \sqrt{2} + 4 \sqrt{2}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
| ⑤ | Multiply both numerator and denominator by -1. |
| ⑥ | Remove 1 from denominator. |
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