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