Answer
$$\frac{5+\sqrt{2}}{5-\sqrt{2}}=\frac{27+10\sqrt{2}}{23}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5+\sqrt{2}}{5-\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5+\sqrt{2}}{5-\sqrt{2}}\frac{5+\sqrt{2}}{5+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{25+5\sqrt{2}+5\sqrt{2}+2}{25+5\sqrt{2}-5\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{27+10\sqrt{2}}{23}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 5 + \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 5 + \sqrt{2}\right) } \cdot \left( 5 + \sqrt{2}\right) = \color{blue}{5} \cdot5+\color{blue}{5} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot5+\color{blue}{ \sqrt{2}} \cdot \sqrt{2} = \\ = 25 + 5 \sqrt{2} + 5 \sqrt{2} + 2 $$ Simplify denominator. $$ \color{blue}{ \left( 5- \sqrt{2}\right) } \cdot \left( 5 + \sqrt{2}\right) = \color{blue}{5} \cdot5+\color{blue}{5} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot5\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 25 + 5 \sqrt{2}- 5 \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
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