Answer
$$\frac{2\sqrt{2}-1}{1+2\sqrt{2}}=\frac{-4\sqrt{2}+9}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{2}-1}{1+2\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{2}-1}{1+2\sqrt{2}}\frac{1-2\sqrt{2}}{1-2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{2}-8-1+2\sqrt{2}}{1-2\sqrt{2}+2\sqrt{2}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4\sqrt{2}-9}{-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-4\sqrt{2}+9}{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 1- 2 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 \sqrt{2}-1\right) } \cdot \left( 1- 2 \sqrt{2}\right) = \color{blue}{ 2 \sqrt{2}} \cdot1+\color{blue}{ 2 \sqrt{2}} \cdot- 2 \sqrt{2}\color{blue}{-1} \cdot1\color{blue}{-1} \cdot- 2 \sqrt{2} = \\ = 2 \sqrt{2}-8-1 + 2 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 1 + 2 \sqrt{2}\right) } \cdot \left( 1- 2 \sqrt{2}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot- 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot1+\color{blue}{ 2 \sqrt{2}} \cdot- 2 \sqrt{2} = \\ = 1- 2 \sqrt{2} + 2 \sqrt{2}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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