Answer
$$\frac{7-\sqrt{3}}{2+\sqrt{3}}=17-9\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7-\sqrt{3}}{2+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7-\sqrt{3}}{2+\sqrt{3}}\frac{2-\sqrt{3}}{2-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{14-7\sqrt{3}-2\sqrt{3}+3}{4-2\sqrt{3}+2\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{17-9\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}17-9\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 7- \sqrt{3}\right) } \cdot \left( 2- \sqrt{3}\right) = \color{blue}{7} \cdot2+\color{blue}{7} \cdot- \sqrt{3}\color{blue}{- \sqrt{3}} \cdot2\color{blue}{- \sqrt{3}} \cdot- \sqrt{3} = \\ = 14- 7 \sqrt{3}- 2 \sqrt{3} + 3 $$ Simplify denominator. $$ \color{blue}{ \left( 2 + \sqrt{3}\right) } \cdot \left( 2- \sqrt{3}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot2+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 4- 2 \sqrt{3} + 2 \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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