Answer
$$\frac{\sqrt{3}-3}{\sqrt{3}+5}=\frac{-9+4\sqrt{3}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}-3}{\sqrt{3}+5}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}-3}{\sqrt{3}+5}\frac{\sqrt{3}-5}{\sqrt{3}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3-5\sqrt{3}-3\sqrt{3}+15}{3-5\sqrt{3}+5\sqrt{3}-25} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{18-8\sqrt{3}}{-22} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{9-4\sqrt{3}}{-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-9+4\sqrt{3}}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}-5} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3}-3\right) } \cdot \left( \sqrt{3}-5\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot-5\color{blue}{-3} \cdot \sqrt{3}\color{blue}{-3} \cdot-5 = \\ = 3- 5 \sqrt{3}- 3 \sqrt{3} + 15 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3} + 5\right) } \cdot \left( \sqrt{3}-5\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot-5+\color{blue}{5} \cdot \sqrt{3}+\color{blue}{5} \cdot-5 = \\ = 3- 5 \sqrt{3} + 5 \sqrt{3}-25 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Multiply both numerator and denominator by -1. |
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