$$\left({\frac{{2+\sqrt 3}}{{2-\sqrt3}}+ \frac{{2 - \sqrt 3}}{{2 + \sqrt 3}} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)$$      Simplifies to :

Solution(By Examveda Team)

$$\left( {\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} + \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)$$
$$ = \left\{ {\frac{{{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2}}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}} \right\}$$
$$ = \left\{ {\frac{{4 + 3 + 4\sqrt 3 + 4 + 3 - 4\sqrt 3 }}{{4 - 3}} + \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{3 - 1}}} \right\}$$
$$\eqalign{ & = \left\{ {14 + \frac{{3 + 1 - 2\sqrt 3 }}{2}} \right\} \cr & = 14 + \frac{{2\left( {2 - \sqrt 3 } \right)}}{2} \cr & = 14 + 2 - \sqrt 3 \cr & = 16 - \sqrt 3 \cr} $$