Question
Answer
$$\dfrac{\sqrt{3}+1}{7-2\sqrt{3}}=\dfrac{9\sqrt{3}+13}{37}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}+1}{7-2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}+1}{7-2\sqrt{3}}\frac{7+2\sqrt{3}}{7+2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7\sqrt{3}+6+7+2\sqrt{3}}{49+14\sqrt{3}-14\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{9\sqrt{3}+13}{37}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7 + 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3} + 1\right) } \cdot \left( 7 + 2 \sqrt{3}\right) = \color{blue}{ \sqrt{3}} \cdot7+\color{blue}{ \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{1} \cdot7+\color{blue}{1} \cdot 2 \sqrt{3} = \\ = 7 \sqrt{3} + 6 + 7 + 2 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 7- 2 \sqrt{3}\right) } \cdot \left( 7 + 2 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot 2 \sqrt{3}\color{blue}{- 2 \sqrt{3}} \cdot7\color{blue}{- 2 \sqrt{3}} \cdot 2 \sqrt{3} = \\ = 49 + 14 \sqrt{3}- 14 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
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