Answer
$$\frac{8+\sqrt{7}}{\sqrt{3}+6}=\frac{-8\sqrt{3}+48-\sqrt{21}+6\sqrt{7}}{33}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{8+\sqrt{7}}{\sqrt{3}+6}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{8+\sqrt{7}}{\sqrt{3}+6}\frac{\sqrt{3}-6}{\sqrt{3}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8\sqrt{3}-48+\sqrt{21}-6\sqrt{7}}{3-6\sqrt{3}+6\sqrt{3}-36} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8\sqrt{3}-48+\sqrt{21}-6\sqrt{7}}{-33} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-8\sqrt{3}+48-\sqrt{21}+6\sqrt{7}}{33}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}-6} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 8 + \sqrt{7}\right) } \cdot \left( \sqrt{3}-6\right) = \color{blue}{8} \cdot \sqrt{3}+\color{blue}{8} \cdot-6+\color{blue}{ \sqrt{7}} \cdot \sqrt{3}+\color{blue}{ \sqrt{7}} \cdot-6 = \\ = 8 \sqrt{3}-48 + \sqrt{21}- 6 \sqrt{7} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3} + 6\right) } \cdot \left( \sqrt{3}-6\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot-6+\color{blue}{6} \cdot \sqrt{3}+\color{blue}{6} \cdot-6 = \\ = 3- 6 \sqrt{3} + 6 \sqrt{3}-36 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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