TH1: \(x\ge-1\) pt tương đương:
\(\sqrt{2x^2-2x+1}+\sqrt{2}\left(x+1\right)=2\)
\(\Leftrightarrow\sqrt{2x^2-2x+1}=\sqrt{2}\left(\sqrt{2}-1-x\right)\) (\(x\le\sqrt{2}-1\))
\(\Leftrightarrow2x^2-2x+1=2\left(x^2-2\left(\sqrt{2}-1\right)x+3-2\sqrt{2}\right)\)
\(\Leftrightarrow-2x+1=-4\left(\sqrt{2}-1\right)x+6-4\sqrt{2}\)
\(\Rightarrow\left(4\sqrt{2}-6\right)x=5-4\sqrt{2}\Rightarrow x=\frac{5-4\sqrt{2}}{4\sqrt{2}-6}>\sqrt{2}-1\left(l\right)\)
TH2: \(x< -1\) pt tương đương:
\(\sqrt{2x^2-2x+1}-\sqrt{2}\left(x+1\right)=2\)
\(\Leftrightarrow\sqrt{2x^2-2x+1}=\sqrt{2}\left(x+\sqrt{2}+1\right)\) (\(x\ge-\sqrt{2}-1\)
\(\Leftrightarrow2x^2-2x+1=2x^2+\left(4\sqrt{2}+4\right)x+6+4\sqrt{2}\)
\(\Leftrightarrow\left(4\sqrt{2}+6\right)x=-5-4\sqrt{2}\)
\(\Rightarrow x=\frac{-5-4\sqrt{2}}{6+4\sqrt{2}}=\frac{1-2\sqrt{2}}{2}\) (thỏa mãn)
