$$2 + \frac{6}{{\sqrt 3 }} + \frac{1}{{2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 - 2}}$$     equals to

Solution(By Examveda Team)

$$\eqalign{ & 2 + \frac{6}{{\sqrt 3 }} + \frac{1}{{2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 - 2}} \cr & \Rightarrow 2 + \frac{{2 \times 3\sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} + \frac{1}{{2 + \sqrt 3 }} - \frac{1}{{2 - \sqrt 3 }} \cr & \Rightarrow 2 + 2\sqrt 3 + \left( {\frac{{\left( {2 - \sqrt 3 } \right) - \left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}} \right) \cr & \Rightarrow 2 + 2\sqrt 3 + \left( {\frac{{2 - \sqrt 3 - 2 - \sqrt 3 }}{{4 - 3}}} \right) \cr & \Rightarrow 2 + 2\sqrt 3 - 2\sqrt 3 \cr & \Rightarrow 2 \cr} $$