$x.(x+7)=0$
$⇒$ \(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\)
Vậy $x$ $∈$ `{0;-7}`
$(x+12).(x-3)=0$
$⇒$ \(\left[ \begin{array}{l}x+12=0\\x-3=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=-12\\x=3\end{array} \right.\)
Vậy $x$ $∈$ `{-12;3}`
$(-x+5).(3-x)=0$
$⇒$ \(\left[ \begin{array}{l}-x+5=0\\3-x=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\)
Vậy $x$ $∈$ `{5;3}`
$x.(2+x).(7-x)=0$
$⇒$ \(\left[ \begin{array}{l}x=0\\2+x=0\\7-x=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=0\\x=-2\\x=7\end{array} \right.\)
Vậy $x$ $∈$ `{0;-2;7}`
$(x-1).(x+2).(-x-3)=0$
$⇒$ \(\left[ \begin{array}{l}x-1=0\\x+2=0\\-x-3=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=1\\x=-2\\x=-3\end{array} \right.\)
Vậy $x$ $∈$ `{1;-2;-3}`
