If $$\sqrt 3 + \frac{1}{{\sqrt 3 }}{\text{,}}$$   then the value of $$\left( {x - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)$$ $$\left( {x - \frac{1}{{x - \frac{{2\sqrt 3 }}{3}}}} \right)$$   is?

Solution (By Examveda Team)

$$\eqalign{ & \because x = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & \Rightarrow x = \frac{{3 + 1}}{{\sqrt 3 }} \cr & \Rightarrow x = \frac{4}{{\sqrt 3 }} \cr & {\text{So, }}\left( {x - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)\left( {x - \frac{1}{{x - \frac{{2\sqrt 3 }}{3}}}} \right) \cr & = \left( {\frac{4}{{\sqrt 3 }} - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)\left( {\frac{4}{{\sqrt 3 }} - \frac{1}{{\frac{4}{{\sqrt 3 }} - \frac{2}{{\sqrt 3 }}}}} \right) \cr & = \left( {\frac{{4\sqrt {42} - \sqrt {126} \times \sqrt 3 }}{{\sqrt {3 \times } \sqrt {42} }}} \right)\left( {\frac{4}{{\sqrt 3 }} - \frac{1}{{\frac{2}{{\sqrt 3 }}}}} \right) \cr & = \frac{{4\sqrt {42} - 3\sqrt {42} }}{{\sqrt {3 \times } \sqrt {42} }}\left( {\frac{4}{{\sqrt 3 }} - \frac{{\sqrt 3 }}{2}} \right) \cr & = \left( {\frac{{\sqrt {42} }}{{\sqrt {3 \times } \sqrt {42} }}} \right)\left( {\frac{{8 - 3}}{{2\sqrt 3 }}} \right) \cr & = \frac{1}{{\sqrt 3 }} \times \frac{5}{{2\sqrt 3 }} \cr & = \frac{5}{6} \cr} $$