\(\sqrt{2x-1}=\sqrt{2}+1\)
\(\Leftrightarrow2x-1=\left(\sqrt{2}+1\right)^2\) điều kiện \(x\ge\frac{1}{2}\)
\(\Leftrightarrow2x-1=3+2\sqrt{2}\)
\(\Leftrightarrow x=\frac{4+2\sqrt{2}}{2}\)
\(\Leftrightarrow x=2+\sqrt{2}\) (TM)
Vậy......
mình nghĩ vậy :)