a)Dk: \(\sqrt{2}x+2\ge0\Rightarrow x\ge-\sqrt{2}\)
\(\sqrt{2}x+2=y\Rightarrow y\ge0\)
\(x=\frac{\left(y-2\right)}{\sqrt{2}};x^2=\frac{y^2-2y+4}{2}\)
\(\Leftrightarrow\sqrt{2y}=\left(\frac{y^2-2y+4}{2}\right)-2\Leftrightarrow y^2-2y-2\sqrt{2y}=0\)\(\Leftrightarrow\left(1\right)y=0=>x=-\sqrt{2}\)