ĐK:\(x\ge\dfrac{5}{3}\)
\(\sqrt{x^2-x}=\sqrt{3x-5}\Leftrightarrow\left(\sqrt{x^2-x}\right)^2=\left(\sqrt{3x-5}\right)^2\Leftrightarrow x^2-x=3x-5\Leftrightarrow x^2-4x+5=0\Leftrightarrow x^2-4x+4+1=0\Leftrightarrow\left(x-1\right)^2+1=0\)
Vì \(\left(x-1\right)^2\ge0\Leftrightarrow\left(x-1\right)^2+1\ge0\)
Vậy phương trình vô nghiệm