ĐK:....
\(\sqrt{x\left(x-1\right)}+\sqrt{x\left(2x-1\right)}=x\)
\(\Leftrightarrow\sqrt{x\left(x-1\right)}-x=-\sqrt{x\left(2x-1\right)}\)
\(\Leftrightarrow x\left(x-1\right)+x^2-2x\sqrt{x\left(x-1\right)}=x\left(2x-1\right)\)
\(\Leftrightarrow x^2-x+x^2-2x\sqrt{x\left(x-1\right)}=2x^2-x\)
\(\Leftrightarrow2x\sqrt{x\left(x-1\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x\left(x-1\right)}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0\\x=1\end{matrix}\right.\)( thỏa )
Vậy....