Answer
$$\frac{5+\sqrt{2}}{3-\sqrt{2}}=\frac{17+8\sqrt{2}}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5+\sqrt{2}}{3-\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5+\sqrt{2}}{3-\sqrt{2}}\frac{3+\sqrt{2}}{3+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{15+5\sqrt{2}+3\sqrt{2}+2}{9+3\sqrt{2}-3\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{17+8\sqrt{2}}{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 + \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 5 + \sqrt{2}\right) } \cdot \left( 3 + \sqrt{2}\right) = \color{blue}{5} \cdot3+\color{blue}{5} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot3+\color{blue}{ \sqrt{2}} \cdot \sqrt{2} = \\ = 15 + 5 \sqrt{2} + 3 \sqrt{2} + 2 $$ Simplify denominator. $$ \color{blue}{ \left( 3- \sqrt{2}\right) } \cdot \left( 3 + \sqrt{2}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot3\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 9 + 3 \sqrt{2}- 3 \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
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