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