`a) -12(x-5)+7(3-x)=-4`
`<=> -12x+60+21-7x=-4`
`<=> -12x-7x=-4-60-21`
`<=> -19x=-85`
`<=> x=85/19`
`b) (x-2)(x+4)=0`
\(⇔\left[ \begin{array}{l}x-2=0\\x+4=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0+2\\x=0-4\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2\\x=-4\end{array} \right.\)
- Vậy `x in {2;-4}`
`c) (x-2)(x+15)=0`
\(⇔\left[ \begin{array}{l}x-2=0\\x+15=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0+2\\x=0-15\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2\\x=-15\end{array} \right.\)
- Vậy `x in {2;-15}`
`d) (7-x)(x+19)=0`
\(⇔\left[ \begin{array}{l}7-x=0\\x+19=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}-x=0-7\\x=0-19\end{array} \right.\)
\(⇔\left[ \begin{array}{l}-x=-7\\x=-19\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=7\\x=-19\end{array} \right.\)
- Vậy `x in {7;-19}`
