ĐK: \(x\ge\dfrac{1}{2}\)
Ta có: \(2+\sqrt{2x-1}=x\)
\(\Leftrightarrow\sqrt{2x-1}=x-2\) ĐK: x≥2
\(\Leftrightarrow2x-1=x^2-4x+4\)
\(\Leftrightarrow x^2-6x+5=0\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=5\left(tm\right)\end{matrix}\right.\)