smc

Let $x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$ and $y=\sqrt{3-2\sqrt{2}}.$ Note that $$\begin{array}{rl}{x}^2& =\frac{{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}^2}{{\left(\sqrt{\sqrt{5}+1}\right)}^2}\\ & =\frac{\left(\sqrt{5}+2\right)+2\cdot \left(\sqrt{\sqrt{5}+2}\cdot \sqrt{\sqrt{5}-2}\right)+\left(\sqrt{5}-2\right)}{\sqrt{5}+1}\\ & =\frac{\left(\sqrt{5}+2\right)+2+\left(\sqrt{5}-2\right)}{\sqrt{5}+1}\\ & =\frac{2\sqrt{5}+2}{\sqrt{5}+1}\\ & =2.\end{array}$$ Since $x>0,$ we have $x=\sqrt{2}.$ On the other hand, note that $$\begin{array}{rl}{y}^2& =3-2\sqrt{2}\\ & =2-2\sqrt{2}+1\\ & ={\left(\sqrt{2}-1\right)}^2.\end{array}$$ Since $y>0,$ we have $y=\sqrt{2}-1.$ Finally, the answer is $$N=x-y=\boxed{1}.$$