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