Question
Answer
$$\dfrac{2}{3-2\sqrt{3}}=-\dfrac{6+4\sqrt{3}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{3-2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{3-2\sqrt{3}}\frac{3+2\sqrt{3}}{3+2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6+4\sqrt{3}}{9+6\sqrt{3}-6\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6+4\sqrt{3}}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{6+4\sqrt{3}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 + 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( 3 + 2 \sqrt{3}\right) = \color{blue}{2} \cdot3+\color{blue}{2} \cdot 2 \sqrt{3} = \\ = 6 + 4 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 3- 2 \sqrt{3}\right) } \cdot \left( 3 + 2 \sqrt{3}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot 2 \sqrt{3}\color{blue}{- 2 \sqrt{3}} \cdot3\color{blue}{- 2 \sqrt{3}} \cdot 2 \sqrt{3} = \\ = 9 + 6 \sqrt{3}- 6 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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