1) \(=\dfrac{\sqrt{\left(3-2\sqrt{2}\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{3-2\sqrt{2}}{\sqrt{2}-1}=\dfrac{\left(3-2\sqrt{2}\right)\left(\sqrt{2}+1\right)}{2-1}=-1+\sqrt{2}\)
2) \(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\sqrt{3}+1=2\)
3) \(=2-\sqrt{3}-2\sqrt{3}+3=5-3\sqrt{3}\)
4) \(=\left[1+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\left[\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-1\right]=\left(1+\sqrt{3}\right)\left(\sqrt{3}-1\right)=3-1=2\)
\(\dfrac{\sqrt{\left(2\sqrt{2}-3\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{3-2\sqrt{2}}{\sqrt{2}-1}=\dfrac{\left(3-2\sqrt{2}\right)\left(\sqrt{2}+1\right)}{2-1}=\sqrt{2}-1\)
\(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\sqrt{3}+1=2\)
\(\sqrt{3}-2-2\sqrt{3}-3=-5-\sqrt{3}\)
\(\left(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)^2-1=\dfrac{6-3\sqrt{3}}{2}-1=\dfrac{4-3\sqrt{3}}{2}\)
