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

$$\eqalign{ & {\text{ }}x = \sqrt 3 + \sqrt 2 \cr & \frac{1}{x} = \frac{1}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \frac{1}{x} = \sqrt 3 - \sqrt 2 \cr & \therefore x + \frac{1}{x} \cr & = \sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 \cr & = 2\sqrt 3 \cr} $$