If $$x = \sqrt {1 + \frac{{\sqrt 3 }}{2}} - \sqrt {1 - \frac{{\sqrt 3 }}{2}} ,$$      then the value of $$\frac{{\sqrt 2 - x}}{{\sqrt 2 + x}}$$  will be closest to:

Solution (By Examveda Team)

$$\eqalign{ & x = \sqrt {1 + \frac{{\sqrt 3 }}{2}} - \sqrt {1 - \frac{{\sqrt 3 }}{2}} \cr & x = \sqrt {\frac{{2 + \sqrt 3 }}{2}} - \sqrt {\frac{{2 - \sqrt 3 }}{2}} \cr & {x^2} = \frac{{2 + \sqrt 3 }}{2} + \frac{{2 - \sqrt 3 }}{2} - \frac{2}{2}\sqrt {4 - 3} \cr & {x^2} = 1 + 1 - 1 \cr & {x^2} = 1 \cr & x = \pm 1 \cr & {\text{At }}x = 1 \cr & \frac{{\sqrt 2 - x}}{{\sqrt 2 + x}} = \frac{{1.414 - 1}}{{1.414 + 1}} = 0.17 \cr} $$