a, Ta có: (x-1)(x+2)<0 => x-1 và x+2 trái dấu. Mà x-1 \(\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x+2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 1\\x>-2\end{matrix}\right.\Rightarrow-2< x< 1\)
Vậy -2 b, Ta có: (x-2)(3-x)>0 => x-2 và 3-x cùng dấu
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\3-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\3-x< 0\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x>3\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>2\\x< 3\end{matrix}\right.\) (do \(\left\{{}\begin{matrix}x< 2\\x>3\end{matrix}\right.\) không thỏa mãn) => 2 Vậy 2
