Question
Answer
$$\dfrac{6-2\sqrt{3}}{6+2\sqrt{3}}=2-\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6-2\sqrt{3}}{6+2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6-2\sqrt{3}}{6+2\sqrt{3}}\frac{6-2\sqrt{3}}{6-2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{36-12\sqrt{3}-12\sqrt{3}+12}{36-12\sqrt{3}+12\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{48-24\sqrt{3}}{24} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2-\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}2-\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6- 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 6- 2 \sqrt{3}\right) } \cdot \left( 6- 2 \sqrt{3}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- 2 \sqrt{3}\color{blue}{- 2 \sqrt{3}} \cdot6\color{blue}{- 2 \sqrt{3}} \cdot- 2 \sqrt{3} = \\ = 36- 12 \sqrt{3}- 12 \sqrt{3} + 12 $$ 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 24. |
| ⑤ | Remove 1 from denominator. |
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