`1)``3x+|x-1|=7`
`TH1:x\geq1`
`->3x+x-1=7`
`->4x=8`
`->x=2``(nhận)`
`TH2:x<1`
`->3x+1-x=7`
`->2x=6`
`->x=3(loại)`
Vậy `S={2}`
`2)``(x-7).(x-2)<0`
`TH1:`
`->`\(\left[ \begin{array}{l}x-7<0\\x-2>0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x<7\\x>2\end{array} \right.\)
`TH2:`
`->`\(\left[ \begin{array}{l}x-7>0\\x-2<0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x>7\\x<2\end{array} \right.\)
`3)``(x-7).(x+11)>0`
`TH1:`
`->`\(\left[ \begin{array}{l}x-7>0\\x+11>0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x>7\\x>-11\end{array} \right.\)
`TH2:`
`->`\(\left[ \begin{array}{l}x-7<0\\x+11<0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x<7\\x<-11\end{array} \right.\)
