​$$\text{Simplify } \left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)^2.$$​​​

Correct option is C

Given:

​​$$\left( \frac{2+\sqrt{3}}{2-\sqrt{3}} \right)^2$$

Solution:

$$\left( \frac{2+\sqrt{3}}{2-\sqrt{3}} \right)^2\\ \ \\ = \left( \frac{2+\sqrt{3}}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}} \right)^2 \\ \ \\ = \left( \frac{(2+\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})} \right)^2 \\ \ \\ = \left( \frac{4+4\sqrt{3}+3}{4-3} \right)^2 \\ \ \\ = \left( \frac{7+4\sqrt{3}}{1} \right)^2 = \left( 7+4\sqrt{3} \right)^2 \\ \ \\ = 49+56\sqrt{3}+48\\ \ \\ = 97+56\sqrt{3}$$

Option (c) is right answer.