Answer
$$\frac{1-\sqrt{3}}{1+\sqrt{3}}=-2+\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1-\sqrt{3}}{1+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1-\sqrt{3}}{1+\sqrt{3}}\frac{1-\sqrt{3}}{1-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1-\sqrt{3}-\sqrt{3}+3}{1-\sqrt{3}+\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4-2\sqrt{3}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2-\sqrt{3}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-2+\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-2+\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 1- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 1- \sqrt{3}\right) } \cdot \left( 1- \sqrt{3}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot- \sqrt{3}\color{blue}{- \sqrt{3}} \cdot1\color{blue}{- \sqrt{3}} \cdot- \sqrt{3} = \\ = 1- \sqrt{3}- \sqrt{3} + 3 $$ Simplify denominator. $$ \color{blue}{ \left( 1 + \sqrt{3}\right) } \cdot \left( 1- \sqrt{3}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot1+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 1- \sqrt{3} + \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Multiply both numerator and denominator by -1. |
| ⑥ | Remove 1 from denominator. |
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