Tìm x,biết: a) (3-x)(x+2)>0 b) (x-2)(x+2020)>0

` a) ` ` (3 – x)(x + 2) > 0 `

` => ` \(\left[ \begin{array}{l}\left \{ {{3-x>0} \atop {x+2>0}} \right. \\\left \{ {{3-x<0} \atop {x+2<0}} \right. \end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}\left \{ {{x>3} \atop {x>-2}} \right. \\\left \{ {{x<3} \atop {x<-2}} \right. \end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}x>3\\x<-2\end{array} \right.\) 

` b) ` ` (x – 2)(x + 2020) > 0 `

` => ` \(\left[ \begin{array}{l}\left \{ {{x-2>0} \atop {x+2020>0}} \right. \\\left \{ {{x-2<0} \atop {x+2020<0}} \right. \end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}\left \{ {{x>2} \atop {x>-2020}} \right. \\\left \{ {{x<2} \atop {x<-2020}} \right. \end{array} \right.\)

` => ` \(\left[ \begin{array}{l}x>2\\x<-2020\end{array} \right.\)