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