Answer
$$\frac{1}{2\sqrt{3}-\sqrt{5}}=\frac{2\sqrt{3}+\sqrt{5}}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{2\sqrt{3}-\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{2\sqrt{3}-\sqrt{5}}\frac{2\sqrt{3}+\sqrt{5}}{2\sqrt{3}+\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{3}+\sqrt{5}}{12+2\sqrt{15}-2\sqrt{15}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{3}+\sqrt{5}}{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{3} + \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 2 \sqrt{3} + \sqrt{5}\right) = \color{blue}{1} \cdot 2 \sqrt{3}+\color{blue}{1} \cdot \sqrt{5} = \\ = 2 \sqrt{3} + \sqrt{5} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{3}- \sqrt{5}\right) } \cdot \left( 2 \sqrt{3} + \sqrt{5}\right) = \color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot \sqrt{5}\color{blue}{- \sqrt{5}} \cdot 2 \sqrt{3}\color{blue}{- \sqrt{5}} \cdot \sqrt{5} = \\ = 12 + 2 \sqrt{15}- 2 \sqrt{15}-5 $$ |
| ③ | Simplify numerator and denominator |
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