Answer
$$\frac{2}{7+3\sqrt{3}}=\frac{7-3\sqrt{3}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{7+3\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{7+3\sqrt{3}}\frac{7-3\sqrt{3}}{7-3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{14-6\sqrt{3}}{49-21\sqrt{3}+21\sqrt{3}-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{14-6\sqrt{3}}{22} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{7-3\sqrt{3}}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7- 3 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( 7- 3 \sqrt{3}\right) = \color{blue}{2} \cdot7+\color{blue}{2} \cdot- 3 \sqrt{3} = \\ = 14- 6 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 7 + 3 \sqrt{3}\right) } \cdot \left( 7- 3 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot7+\color{blue}{ 3 \sqrt{3}} \cdot- 3 \sqrt{3} = \\ = 49- 21 \sqrt{3} + 21 \sqrt{3}-27 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
This page was created using
Rationalize Denominator Calculator