a)
(x-1)(x-2) > 0 ⇔ \(\left[ \begin{array}{l} \left \{ {{x-1>0} \atop {x-2>0}} \right. \\\left \{ {{x-1<0} \atop {x-2<0}} \right. \end{array} \right.\)
⇔\(\left[ \begin{array}{l}\left \{ {{x>1} \atop {x>2}} \right.\\ \left \{ {{x<1} \atop {x<2}} \right.\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x>2\\x<1\end{array} \right.\)
Vậy x > 2 hoặc x < 1 thì (x-1).(x-2) > 0
b)
(x+3) . (x-2) < 0 ⇔ \(\left[ \begin{array}{l}\left \{ {{x+3<0} \atop {x-2>0}} \right.\\\left \{ {{x+3>0} \atop {x-2<0}} \right. \end{array} \right.\)
⇔\(\left[ \begin{array}{l}\left \{ {{x<-3} \atop {x>2}}(Vô lý) \right.\\\left \{ {{x>-3} \atop {x<2}} \right. \end{array} \right.\)
⇔$-3<x<2$
Vậy $-3<x<2$ thì (x+3) . (x-2) < 0
