Answer
$$\frac{3+\sqrt{3}}{6-2\sqrt{3}}=\frac{2+\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3+\sqrt{3}}{6-2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+\sqrt{3}}{6-2\sqrt{3}}\frac{6+2\sqrt{3}}{6+2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{18+6\sqrt{3}+6\sqrt{3}+6}{36+12\sqrt{3}-12\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{24+12\sqrt{3}}{24} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2+\sqrt{3}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6 + 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 + \sqrt{3}\right) } \cdot \left( 6 + 2 \sqrt{3}\right) = \color{blue}{3} \cdot6+\color{blue}{3} \cdot 2 \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot6+\color{blue}{ \sqrt{3}} \cdot 2 \sqrt{3} = \\ = 18 + 6 \sqrt{3} + 6 \sqrt{3} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( 6- 2 \sqrt{3}\right) } \cdot \left( 6 + 2 \sqrt{3}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot 2 \sqrt{3}\color{blue}{- 2 \sqrt{3}} \cdot6\color{blue}{- 2 \sqrt{3}} \cdot 2 \sqrt{3} = \\ = 36 + 12 \sqrt{3}- 12 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 12. |
This page was created using
Rationalize Denominator Calculator