a)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 hoặc x=-7
b)(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 hoặc x=3
c)(-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 hoặc 3
d)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= \(\left[ \begin{array}{l}x=0\\x=-2\\x=7\end{array} \right.\)
e)(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=\(\left[ \begin{array}{l}x=1\\x=-2\\x=-3\end{array} \right.\)
