Answer
$$\frac{\sqrt{3}+2\sqrt{5}}{\sqrt{3}+\sqrt{5}}=\frac{7-\sqrt{15}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}+2\sqrt{5}}{\sqrt{3}+\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}+2\sqrt{5}}{\sqrt{3}+\sqrt{5}}\frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3-\sqrt{15}+2\sqrt{15}-10}{3-\sqrt{15}+\sqrt{15}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-7+\sqrt{15}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{7-\sqrt{15}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}- \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3} + 2 \sqrt{5}\right) } \cdot \left( \sqrt{3}- \sqrt{5}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{5}+\color{blue}{ 2 \sqrt{5}} \cdot \sqrt{3}+\color{blue}{ 2 \sqrt{5}} \cdot- \sqrt{5} = \\ = 3- \sqrt{15} + 2 \sqrt{15}-10 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3} + \sqrt{5}\right) } \cdot \left( \sqrt{3}- \sqrt{5}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot \sqrt{3}+\color{blue}{ \sqrt{5}} \cdot- \sqrt{5} = \\ = 3- \sqrt{15} + \sqrt{15}-5 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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