$x=\frac{2}{\sqrt{2}}+1=\frac{2\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}+1=\frac{2\sqrt{2}}{2}+1=\sqrt{2}+1$

Sposób I: $x^2-2xy+y^2={\left(\sqrt{2}+1\right)}^2-2\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)+{\left(\sqrt{2}-1\right)}^2=$

                 $=2+2\sqrt{2}+1-2\left(2-1\right)+2-2\sqrt{2}+1=$

                 $=3+2\sqrt{2}-2\cdot 1+3-2\sqrt{2}=3-2+3=4$

Sposób II: $x^2-2xy+y^2={\left(x-y\right)}^2={\left(\sqrt{2}+1-\left(\sqrt{2}-1\right)\right)}^2=$

                  $={\left(\sqrt{2}+1-\sqrt{2}+1\right)}^2=2^2=4$