*Lời giải :
`(x + 5) (x - 2) < 0`
`⇔` \(\left\{ \begin{array}{l}x+5<0\\x-2>0\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x<-5\\x>2\end{array} \right.\) `⇔ 2 < x < -5` (Vô lí `->` loại)
`⇔` \(\left\{ \begin{array}{l}x+5>0\\x-2<0\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x>-5\\x<2\end{array} \right.\) `⇔ -5 < x < 2` (Nhận)
Ta có : `-5 < x < 2`
`→ x ∈ {-4;-3;-2;-1;0;1}`
Vậy `x ∈ {-4;-3;-2;-1;0;1}`
