Answer
$$\frac{8+\sqrt{3}}{\sqrt{3}-1}=\frac{9\sqrt{3}+11}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{8+\sqrt{3}}{\sqrt{3}-1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{8+\sqrt{3}}{\sqrt{3}-1}\frac{\sqrt{3}+1}{\sqrt{3}+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8\sqrt{3}+8+3+\sqrt{3}}{3+\sqrt{3}-\sqrt{3}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{9\sqrt{3}+11}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3} + 1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 8 + \sqrt{3}\right) } \cdot \left( \sqrt{3} + 1\right) = \color{blue}{8} \cdot \sqrt{3}+\color{blue}{8} \cdot1+\color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot1 = \\ = 8 \sqrt{3} + 8 + 3 + \sqrt{3} $$ 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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