Answer
$$\frac{7}{1+\sqrt{3}}=\frac{-7+7\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7}{1+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7}{1+\sqrt{3}}\frac{1-\sqrt{3}}{1-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7-7\sqrt{3}}{1-\sqrt{3}+\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7-7\sqrt{3}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-7+7\sqrt{3}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 1- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 } \cdot \left( 1- \sqrt{3}\right) = \color{blue}{7} \cdot1+\color{blue}{7} \cdot- \sqrt{3} = \\ = 7- 7 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 1 + \sqrt{3}\right) } \cdot \left( 1- \sqrt{3}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot1+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 1- \sqrt{3} + \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
This page was created using
Rationalize Denominator Calculator