Answer
$$\frac{2\sqrt{3}-\sqrt{5}}{2\sqrt{5}+3\sqrt{3}}=-\sqrt{15}+4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{3}-\sqrt{5}}{2\sqrt{5}+3\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{3}-\sqrt{5}}{2\sqrt{5}+3\sqrt{3}}\frac{2\sqrt{5}-3\sqrt{3}}{2\sqrt{5}-3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4\sqrt{15}-18-10+3\sqrt{15}}{20-6\sqrt{15}+6\sqrt{15}-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7\sqrt{15}-28}{-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{15}-4}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-\sqrt{15}+4}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-\sqrt{15}+4\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{5}- 3 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 \sqrt{3}- \sqrt{5}\right) } \cdot \left( 2 \sqrt{5}- 3 \sqrt{3}\right) = \color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{5}+\color{blue}{ 2 \sqrt{3}} \cdot- 3 \sqrt{3}\color{blue}{- \sqrt{5}} \cdot 2 \sqrt{5}\color{blue}{- \sqrt{5}} \cdot- 3 \sqrt{3} = \\ = 4 \sqrt{15}-18-10 + 3 \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{5} + 3 \sqrt{3}\right) } \cdot \left( 2 \sqrt{5}- 3 \sqrt{3}\right) = \color{blue}{ 2 \sqrt{5}} \cdot 2 \sqrt{5}+\color{blue}{ 2 \sqrt{5}} \cdot- 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot 2 \sqrt{5}+\color{blue}{ 3 \sqrt{3}} \cdot- 3 \sqrt{3} = \\ = 20- 6 \sqrt{15} + 6 \sqrt{15}-27 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 7. |
| ⑤ | Multiply both numerator and denominator by -1. |
| ⑥ | Remove 1 from denominator. |
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