\[\begin{array}{l}\left| {x - 1} \right| > 3 \Leftrightarrow \left[ \begin{array}{l}x - 1 > 3\\x - 1 < - 3\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x > 4\\x < - 2\end{array} \right.\\ \Rightarrow A = \left[ { - \infty ; - 2} \right] \cup \left[ {4; + \infty } \right]\\\left| {x + 2} \right| < 5 \Leftrightarrow - 5 < x + 2 < 5\\ \Leftrightarrow - 7 < x < 3\\ \Rightarrow B = \left[ { - 7;3} \right]\\ \Rightarrow A \cap B = \left[ { - 7; - 2} \right]\end{array}\]
Đề bài
Cho \[A = \left\{ {x \in R|\left| {x - 1} \right| > 3} \right\}\] và \[B = \left\{ {x \in R|\left| {x + 2} \right| < 5} \right\}.\]
Tìm \[A \cap B.\]
Lời giải chi tiết
Ta có:
\[\begin{array}{l}\left| {x - 1} \right| > 3 \Leftrightarrow \left[ \begin{array}{l}x - 1 > 3\\x - 1 < - 3\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x > 4\\x < - 2\end{array} \right.\\ \Rightarrow A = \left[ { - \infty ; - 2} \right] \cup \left[ {4; + \infty } \right]\\\left| {x + 2} \right| < 5 \Leftrightarrow - 5 < x + 2 < 5\\ \Leftrightarrow - 7 < x < 3\\ \Rightarrow B = \left[ { - 7;3} \right]\\ \Rightarrow A \cap B = \left[ { - 7; - 2} \right]\end{array}\]
Vậy \[A \cap B = \left[ { - 7; - 2} \right]\]