Đặt \(\sqrt{2x-1}=a\ge0\)
Ta có \(2011x^2-a^2=2010xa\)
\(\Leftrightarrow\left(2010x^2-2010xa\right)+\left(x^2-a^2\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left(2010x+x+a\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=a\\2011x=-a\left(loai\right)\end{cases}}\)
\(\Leftrightarrow x=1\)