\(^{x^2}\).(x-1)-5x.(1-x)=0
\(\Leftrightarrow\)\(^{x^2}\).(x-1)+5x.(x-1)=0
\(\Leftrightarrow\)(x-1).(\(^{x^2}\)+5x)=0
\(\Leftrightarrow\)(x-1).x.(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-1=0\\x=0\\x+5=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\\x=0\\x=-5\end{matrix}\right.\)
Vậy x\(\in\)\(\left\{-5;0;1\right\}\)