Đáp án:
$ (-\infty;1] \cup \{5\}\cup [9;+\infty).$
Giải thích các bước giải:
$A \cap (B \cup C)=(A \cap B) \cup (A \cap C)$
Chứng minh:
Lấy $x \in A \cap (B \cup C)$
$\Rightarrow \left\{\begin{array}{l} x \in A \\ x \in (B \cup C)\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \in A \\ \left[\begin{array}{l} x \in B \\ x \in C\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x \in A \\ x \in B\end{array} \right.\\\left\{\begin{array}{l} x \in A \\ x \in C\end{array} \right.\end{array} \right.\\ \Leftrightarrow x \in (A \cap B) \cup (A \cap C)\\ \Rightarrow A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)(1)$
Lấy $x \in (A \cap B) \cup (A \cap C)$
$\Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x \in A \\ x \in B\end{array} \right.\\\left\{\begin{array}{l} x \in A \\ x \in C\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \in A \\ \left[\begin{array}{l} x \in B \\ x \in C\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \in A \\ x \in (B \cup C)\end{array} \right.\\ \Leftrightarrow x \in A \cap (B \cup C)\\ \Rightarrow (A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)(2)\\ (1)(2)\Rightarrow A \cap (B \cup C)=(A \cap B) \cup (A \cap C)$
Áp dụng:
$\Big((-\infty;1] \cup [5;+\infty)\Big) \cap \Big((-\infty;5] \cup [9;+\infty)\Big)\\ =\Bigg(\Big((-\infty;1] \cup [5;+\infty)\Big) \cap (-\infty;5] \Bigg) \cup \Bigg(\Big((-\infty;1] \cup [5;+\infty)\Big) \cap [9;+\infty) \Bigg)\\ = (-\infty;1] \cup \{5\}\cup [9;+\infty).$
