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