Question
Answer
$$\dfrac{2+\sqrt{3}}{3+\sqrt{3}}=\dfrac{3+\sqrt{3}}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2+\sqrt{3}}{3+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2+\sqrt{3}}{3+\sqrt{3}}\frac{3-\sqrt{3}}{3-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6-2\sqrt{3}+3\sqrt{3}-3}{9-3\sqrt{3}+3\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3+\sqrt{3}}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 + \sqrt{3}\right) } \cdot \left( 3- \sqrt{3}\right) = \color{blue}{2} \cdot3+\color{blue}{2} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot3+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 6- 2 \sqrt{3} + 3 \sqrt{3}-3 $$ Simplify denominator. $$ \color{blue}{ \left( 3 + \sqrt{3}\right) } \cdot \left( 3- \sqrt{3}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot3+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 9- 3 \sqrt{3} + 3 \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
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