a)
(x-3)(1-x)>0 ⇔ \(\left[ \begin{array}{l}\left \{ {{x-3>0} \atop {1-x>0}} \right. \\\left \{ {{x-3<0} \atop {1-x<0}} \right.\end{array} \right.\)
⇔\(\left[ \begin{array}{l}\left \{ {{x>3} \atop {x<1}}(Vô lý) \right.\\\left \{ {{x<3} \atop {x>1}} \right.\end{array} \right.\)
⇔ $1<x<3$
b)
`x^2+x+1=x^2+x+1/4+3/4=(x+1/2)^2+3/4`
Vì `(x+1/2)^2+3/4≥0∀x` ⇒ `x^2+x+1≥0∀x`
`(x+2)(x-3)(x^2+x+1)>0 `⇔ (x+2)(x-3)>0
⇔\(\left[ \begin{array}{l}\left \{ {{x+2>0} \atop {x-3>0}} \right.\\\left \{ {{x+2<0} \atop {x-3<0}} \right.\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\left \{ {{x>-2} \atop {x>3}} \right.\\\left \{ {{x<-2} \atop {x<3}} \right.\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x>3\\x<-2\end{array} \right.\)
⇔ $x>3$ hoặc $x<-2$
