If $$x = 5 + 2\sqrt 6 {\text{,}}$$    then $$\sqrt x - \frac{1}{{\sqrt x }}$$   = is?

$$\eqalign{ & {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2} \cr & = x + \frac{1}{x} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} \times \frac{{\left( {5 - 2\sqrt 6 } \right)}}{{\left( {5 - 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \left( {5 - 2\sqrt 6 } \right) - 2 \cr & = 10 - 2 \cr & = 8 \cr & \therefore \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right) = \sqrt 8 = 2\sqrt 2 \cr} $$