Question
Answer
$$\dfrac{3}{4-3\sqrt{3}}=-\dfrac{12+9\sqrt{3}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3}{4-3\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3}{4-3\sqrt{3}}\frac{4+3\sqrt{3}}{4+3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12+9\sqrt{3}}{16+12\sqrt{3}-12\sqrt{3}-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12+9\sqrt{3}}{-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{12+9\sqrt{3}}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4 + 3 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 } \cdot \left( 4 + 3 \sqrt{3}\right) = \color{blue}{3} \cdot4+\color{blue}{3} \cdot 3 \sqrt{3} = \\ = 12 + 9 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 4- 3 \sqrt{3}\right) } \cdot \left( 4 + 3 \sqrt{3}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot 3 \sqrt{3}\color{blue}{- 3 \sqrt{3}} \cdot4\color{blue}{- 3 \sqrt{3}} \cdot 3 \sqrt{3} = \\ = 16 + 12 \sqrt{3}- 12 \sqrt{3}-27 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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