Question
Answer
$$\dfrac{4}{\dfrac{1}{3}+\sqrt{3}}=\dfrac{-\dfrac{4}{3}+4\sqrt{3}}{\dfrac{26}{9}}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{\frac{1}{3}+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{\frac{1}{3}+\sqrt{3}}\frac{\frac{1}{3}-\sqrt{3}}{\frac{1}{3}-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\frac{4}{3}-4\sqrt{3}}{\frac{1}{9}+(-\frac{1}{3})\sqrt{3}+\frac{1}{3}\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\frac{4}{3}-4\sqrt{3}}{-\frac{26}{9}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-\frac{4}{3}+4\sqrt{3}}{\frac{26}{9}}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \frac{ 1 }{ 3 }- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( \frac{ 1 }{ 3 }- \sqrt{3}\right) = \color{blue}{4} \cdot\frac{ 1 }{ 3 }+\color{blue}{4} \cdot- \sqrt{3} = \\ = \frac{ 4 }{ 3 }- 4 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \frac{ 1 }{ 3 } + \sqrt{3}\right) } \cdot \left( \frac{ 1 }{ 3 }- \sqrt{3}\right) = \color{blue}{\frac{ 1 }{ 3 }} \cdot\frac{ 1 }{ 3 }+\color{blue}{\frac{ 1 }{ 3 }} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot\frac{ 1 }{ 3 }+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = \frac{ 1 }{ 9 }- \frac{ 1 }{ 3 } \sqrt{ 3 } + \frac{ 1 }{ 3 } \sqrt{ 3 }-3 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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