Trả lời bởi giáo viên

Đáp án đúng: a

\(\begin{array}{l}\left| {x - 2} \right| > x + 1 \Leftrightarrow \left[ \begin{array}{l}x + 1 < 0\\\left\{ \begin{array}{l}x + 1 \ge 0\\{\left( {x - 2} \right)^2} > {\left( {x + 1} \right)^2}\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x <  - 1\\\left\{ \begin{array}{l}x \ge  - 1\\4 - 4x > 2x + 1\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x <  - 1\\\left\{ \begin{array}{l}x \ge  - 1\\6x < 3\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x <  - 1\\\left\{ \begin{array}{l}x \ge  - 1\\x < \dfrac{1}{2}\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x <  - 1\\ - 1 \le x < \dfrac{1}{2}\end{array} \right. \Leftrightarrow x < \dfrac{1}{2}.\end{array}\)

Hướng dẫn giải:

\(\left| A \right| > B \Leftrightarrow \left[ \begin{array}{l}B < 0\\\left\{ \begin{array}{l}B \ge 0\\{A^2} > {B^2}\end{array} \right.\end{array} \right..\)

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