$a) (5-x).(2+x)<0$
TH $1$ : \(\left[ \begin{array}{l}5-x<0\\2+x>0\end{array} \right.\) =>\(\left[ \begin{array}{l}x>5\\x>-2\end{array} \right.\) $=>x>5$
TH $2$: \(\left[ \begin{array}{l}5-x>0\\2+x<0\end{array} \right.\) =>\(\left[ \begin{array}{l}x<5\\x<-2\end{array} \right.\) =>$x<-2$
Vậy với mọi $5<x<-2$ (vô lí)
Phương trình kh thể xác định nghiệm
$b) ,(5-x).(x+2)>0
TH $1$ : \(\left[ \begin{array}{l}5-x<0\\2+x<0\end{array} \right.\) =>\(\left[ \begin{array}{l}x>5\\x<-2\end{array} \right.\) (vô lí)
TH $2$: \(\left[ \begin{array}{l}5-x>0\\2+x>0\end{array} \right.\) =>\(\left[ \begin{array}{l}x<5\\x>-2\end{array} \right.\) =>$-2<x<5$
Vậy với $-2<x<5$ thì $(5-x).(x+2)>0$
