Cho hai tập khác rỗng :$A = \left( {m-1;4} \right],{\rm{ }}B = \left( {-2{\rm{ }};2m + 2} \right)$, với  $m \in \mathbb{R}.$

Xác định $m$  để:$(A \cap B) \subset ( - 1\,;\,\,3)$.

Với $A = \left( {m-1;\,\,4} \right],{\rm{ }}B = \left( {-2;\,\,2m + 2} \right)$ khác tập rỗng.

$\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}m - 1 < 4\\2m + 2 >  - 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m < 5\\m >  - 2\end{array} \right.\\ \Leftrightarrow  - 2 < m < 5\,\,\,\,\,\left( * \right).\\ \Rightarrow A \cap B = \left[ \begin{array}{l}\left( {m - 1;\,\,2m + 2} \right)\\\left( { - 2;\,\,2m + 2} \right)\\\left( {m - 1;\,\,4} \right]\\\left( { - 2;\,\,4} \right]\end{array} \right. \\\left( {A \cap B} \right) \subset \left( { - 1\,;\,\,3} \right)\\ \Leftrightarrow \left( {m - 1;\,\,2m + 2} \right) \subset \left( { - 1\,;\,\,3} \right) \\ \Leftrightarrow \left\{ \begin{array}{l}m - 1 \ge  - 1\\2m + 2 \le 3\end{array} \right. \Leftrightarrow 0 \le m \le \dfrac{1}{2}\,\,\,\,\left( {tm\,\,\,\left( * \right)} \right)\end{array}$