If $$x = \frac{1}{{\left( {\sqrt 2 + 1} \right)}}{\text{,}}$$    the value of x² + 2x - 1 is?

$$\eqalign{ & {\text{Given that,}} \cr & {\text{ }}x = \frac{1}{{\left( {\sqrt 2 + 1} \right)}}{\text{ }} \cr & {\text{Then, }}{x^2} + 2x - 1 \cr & = {x^2} + 2x - 1 + 1 - 1 \cr & = {x^2} + 2x + 1 - 2 \cr & = {\left( {x + 1} \right)^2} - 2 \cr & {\text{Now put the value of }}x \cr & = {\left( {\frac{1}{{\left( {\sqrt 2 + 1} \right)}} + 1} \right)^2} - 2 \cr & = {\left( {\frac{{1 + \sqrt 2 + 1}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\frac{{\sqrt 2 + 2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\frac{{\left( {\sqrt 2 + 1} \right) \times \sqrt2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\sqrt 2 } \right)^2} - 2 \cr & = 2 - 2 \cr & = 0 \cr} $$