Answer
$$\frac{1+\sqrt{2}}{3+\sqrt{5}}=\frac{3-\sqrt{5}+3\sqrt{2}-\sqrt{10}}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1+\sqrt{2}}{3+\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1+\sqrt{2}}{3+\sqrt{5}}\frac{3-\sqrt{5}}{3-\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3-\sqrt{5}+3\sqrt{2}-\sqrt{10}}{9-3\sqrt{5}+3\sqrt{5}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3-\sqrt{5}+3\sqrt{2}-\sqrt{10}}{4}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 1 + \sqrt{2}\right) } \cdot \left( 3- \sqrt{5}\right) = \color{blue}{1} \cdot3+\color{blue}{1} \cdot- \sqrt{5}+\color{blue}{ \sqrt{2}} \cdot3+\color{blue}{ \sqrt{2}} \cdot- \sqrt{5} = \\ = 3- \sqrt{5} + 3 \sqrt{2}- \sqrt{10} $$ Simplify denominator. $$ \color{blue}{ \left( 3 + \sqrt{5}\right) } \cdot \left( 3- \sqrt{5}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot3+\color{blue}{ \sqrt{5}} \cdot- \sqrt{5} = \\ = 9- 3 \sqrt{5} + 3 \sqrt{5}-5 $$ |
| ③ | Simplify numerator and denominator |
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