Ta có: 7x(x-5)-(x-5)=0
⇔(x-5)(7x-1)=0
⇔\(\left[{}\begin{matrix}x-5=0\\7x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\7x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{1}{7}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{1}{7}\right\}\)