Answer
$$\frac{2+\sqrt{3}}{3-\sqrt{5}}=\frac{6+2\sqrt{5}+3\sqrt{3}+\sqrt{15}}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2+\sqrt{3}}{3-\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2+\sqrt{3}}{3-\sqrt{5}}\frac{3+\sqrt{5}}{3+\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6+2\sqrt{5}+3\sqrt{3}+\sqrt{15}}{9+3\sqrt{5}-3\sqrt{5}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6+2\sqrt{5}+3\sqrt{3}+\sqrt{15}}{4}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 + \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 + \sqrt{3}\right) } \cdot \left( 3 + \sqrt{5}\right) = \color{blue}{2} \cdot3+\color{blue}{2} \cdot \sqrt{5}+\color{blue}{ \sqrt{3}} \cdot3+\color{blue}{ \sqrt{3}} \cdot \sqrt{5} = \\ = 6 + 2 \sqrt{5} + 3 \sqrt{3} + \sqrt{15} $$ 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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