Answer
$$\frac{6+\sqrt{2}}{8-\sqrt{2}}=\frac{25+7\sqrt{2}}{31}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6+\sqrt{2}}{8-\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6+\sqrt{2}}{8-\sqrt{2}}\frac{8+\sqrt{2}}{8+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{48+6\sqrt{2}+8\sqrt{2}+2}{64+8\sqrt{2}-8\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{50+14\sqrt{2}}{62} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{25+7\sqrt{2}}{31}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 8 + \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 6 + \sqrt{2}\right) } \cdot \left( 8 + \sqrt{2}\right) = \color{blue}{6} \cdot8+\color{blue}{6} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot8+\color{blue}{ \sqrt{2}} \cdot \sqrt{2} = \\ = 48 + 6 \sqrt{2} + 8 \sqrt{2} + 2 $$ Simplify denominator. $$ \color{blue}{ \left( 8- \sqrt{2}\right) } \cdot \left( 8 + \sqrt{2}\right) = \color{blue}{8} \cdot8+\color{blue}{8} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot8\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 64 + 8 \sqrt{2}- 8 \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
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