Answer
$$\frac{1}{6\sqrt{3}-3}=\frac{2\sqrt{3}+1}{33}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{6\sqrt{3}-3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{6\sqrt{3}-3}\frac{6\sqrt{3}+3}{6\sqrt{3}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{3}+3}{108+18\sqrt{3}-18\sqrt{3}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6\sqrt{3}+3}{99} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2\sqrt{3}+1}{33}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6 \sqrt{3} + 3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 6 \sqrt{3} + 3\right) = \color{blue}{1} \cdot 6 \sqrt{3}+\color{blue}{1} \cdot3 = \\ = 6 \sqrt{3} + 3 $$ Simplify denominator. $$ \color{blue}{ \left( 6 \sqrt{3}-3\right) } \cdot \left( 6 \sqrt{3} + 3\right) = \color{blue}{ 6 \sqrt{3}} \cdot 6 \sqrt{3}+\color{blue}{ 6 \sqrt{3}} \cdot3\color{blue}{-3} \cdot 6 \sqrt{3}\color{blue}{-3} \cdot3 = \\ = 108 + 18 \sqrt{3}- 18 \sqrt{3}-9 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
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