If N = 5 + 2 + 5 - 2 5 + 1 - 3 - 2 2 , then the value of N is = ?

Answer: Option C

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