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