Đặt $A=\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$
$\Rightarrow A^2=\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}}{\sqrt{5}+1}=\dfrac{2\sqrt{5}+2}{\sqrt{5}+1}=\dfrac{2(\sqrt{5}+1)}{\sqrt{5}+1}=2
\\\Rightarrow A=\sqrt{2}
\\\Rightarrow \dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}=\sqrt{2}-\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-\sqrt{2}+1=1$