Question
Answer
$$\dfrac{4}{3+3\sqrt{3}}=\dfrac{-12+12\sqrt{3}}{18}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{3+3\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{3+3\sqrt{3}}\frac{3-3\sqrt{3}}{3-3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12-12\sqrt{3}}{9-9\sqrt{3}+9\sqrt{3}-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12-12\sqrt{3}}{-18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-12+12\sqrt{3}}{18}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- 3 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( 3- 3 \sqrt{3}\right) = \color{blue}{4} \cdot3+\color{blue}{4} \cdot- 3 \sqrt{3} = \\ = 12- 12 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 3 + 3 \sqrt{3}\right) } \cdot \left( 3- 3 \sqrt{3}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot3+\color{blue}{ 3 \sqrt{3}} \cdot- 3 \sqrt{3} = \\ = 9- 9 \sqrt{3} + 9 \sqrt{3}-27 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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