Đáp án:
Giải thích các bước giải:
`1)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}`
`2)(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}`
`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∈{3;5}`
`4)x(x+2)(7-x)=0`
`→` \(\left[ \begin{array}{l}x=0\\x+2=0\\7-x=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x=-2\\x=7\end{array} \right.\)
Vậy `x∈{0;-2;7}`
`5)(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}`
