`5x . (x - 1) = x - 1`
`⇔ 5x (x - 1) - (x - 1) = 0`
` ⇔ (5x - 1) (x - 1) = 0`
⇔ \(\left[ \begin{array}{l}5x-1=0\\x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}5x=1\\x=1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\dfrac{1}{5}\\x=1\end{array} \right.\)
Vậy `x ∈ {`$\dfrac{1}{5}$ `; 1}`
