Answer
$$\frac{4\sqrt{3}}{2\sqrt{3}-3}=\frac{24+12\sqrt{3}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4\sqrt{3}}{2\sqrt{3}-3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4\sqrt{3}}{2\sqrt{3}-3}\frac{2\sqrt{3}+3}{2\sqrt{3}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{24+12\sqrt{3}}{12+6\sqrt{3}-6\sqrt{3}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{24+12\sqrt{3}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{3} + 3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 \sqrt{3} } \cdot \left( 2 \sqrt{3} + 3\right) = \color{blue}{ 4 \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ 4 \sqrt{3}} \cdot3 = \\ = 24 + 12 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{3}-3\right) } \cdot \left( 2 \sqrt{3} + 3\right) = \color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot3\color{blue}{-3} \cdot 2 \sqrt{3}\color{blue}{-3} \cdot3 = \\ = 12 + 6 \sqrt{3}- 6 \sqrt{3}-9 $$ |
| ③ | Simplify numerator and denominator |
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