If N = $$\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} - $$     $$\sqrt {3 - 2\sqrt 2 } {\text{,}}$$   then the value of N is = ?

Solution(By Examveda Team)

$$\eqalign{ & {\text{Let X = }}\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} \cr & {\text{Then,}} \cr & {{\text{X}}^2} = \frac{{{{\left( {\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} } \right)}^2}}}{{{{\left( {\sqrt {\sqrt 5 + 1} } \right)}^2}}} \cr} $$
  $${{\text{X}}^2} = {\text{ }}\frac{{\left( {\sqrt 5 + 2} \right) + \left( {\sqrt 5 - 2} \right) + 2\sqrt {\left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 - 2} \right)} }}{{\left( {\sqrt 5 + 1} \right)}}$$
$$\eqalign{ & {{\text{X}}^2} = \frac{{2\sqrt 5 + 2\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( 2 \right)}^2}} }}{{\sqrt 5 + 1}} \cr & {{\text{X}}^2} = \frac{{2\sqrt 5 + 2}}{{\sqrt 5 + 1}} \cr & {{\text{X}}^2} = \frac{{2\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr & {{\text{X}}^2} = 2 \cr & {\text{X = }}\sqrt 2 \cr & \therefore {\text{N}} = \sqrt 2 - \sqrt {3 - 2\sqrt 2 } \cr & {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {1^2} - 2 \times \sqrt 2 \times } 1 \cr & {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \cr & {\text{N}} = \sqrt 2 - \left( {\sqrt 2 - 1} \right) \cr & {\text{N}} = 1 \cr} $$