ĐK: 2x−1≥0⇔x≥\(\frac{1}{2}\)
\(x^2\)−2x=2\(\sqrt{2x-1}\)
⇔\(x^2\)=\(\sqrt{\left(2x-1\right)^2}+2\sqrt{2x-1}+1\)
⇔\(x^2\)=\(\left(\sqrt{2x-1}+1\right)^2\)
TH1: x=√2x−1+1⇔x=2±√2x=2x-1+1⇔x=2±\(\sqrt{2}\)
TH2: x= -\(\sqrt{2x-1}\)⇔x=±\(\sqrt{2}\)
vậy s = {2+\(\sqrt{2}\) ; \(\sqrt{2}\)}