$ (2x-4) (x + 1) =0$
⇒\(\left[ \begin{array}{l}2x-4=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x=4\\x=-1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
$Vậy$ $x=2 $ $hoặc$ $x=-1$
$3.x(x-1)=0$
⇒\(\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
$Vậy $ $x=0$ $hoặc$ $x=1$
