Question
Answer
$$\dfrac{5+2\sqrt{3}}{2+2\sqrt{3}}=\dfrac{1+3\sqrt{3}}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5+2\sqrt{3}}{2+2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5+2\sqrt{3}}{2+2\sqrt{3}}\frac{2-2\sqrt{3}}{2-2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{10-10\sqrt{3}+4\sqrt{3}-12}{4-4\sqrt{3}+4\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-2-6\sqrt{3}}{-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-1-3\sqrt{3}}{-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{1+3\sqrt{3}}{4}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 5 + 2 \sqrt{3}\right) } \cdot \left( 2- 2 \sqrt{3}\right) = \color{blue}{5} \cdot2+\color{blue}{5} \cdot- 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot2+\color{blue}{ 2 \sqrt{3}} \cdot- 2 \sqrt{3} = \\ = 10- 10 \sqrt{3} + 4 \sqrt{3}-12 $$ Simplify denominator. $$ \color{blue}{ \left( 2 + 2 \sqrt{3}\right) } \cdot \left( 2- 2 \sqrt{3}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot2+\color{blue}{ 2 \sqrt{3}} \cdot- 2 \sqrt{3} = \\ = 4- 4 \sqrt{3} + 4 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Multiply both numerator and denominator by -1. |
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