\(\left| {{x^2} - x} \right| > 2 - x\).

D

- Hướng dẫn giải

Phương pháp giải:

\(\left| {f\left( x \right)} \right| > g\left( x \right) \Leftrightarrow \left[ \begin{array}{l}g\left( x \right) < 0\\\left\{ \begin{array}{l}g\left( x \right) > 0\\{f^2}\left( x \right) > g\left( x \right)\end{array} \right.\end{array} \right..\)

Giải chi tiết:

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

\( \Leftrightarrow \left[ \begin{array}{l}x > 2\\\left\{ \begin{array}{l}x \le 2\\\left[ \begin{array}{l}x > \sqrt 2 \\x <  - \sqrt 2 \end{array} \right.\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x > 2\\\sqrt 2  < x \le 2\\x <  - \sqrt 2 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x > \sqrt 2 \\x <  - \sqrt 2 \end{array} \right..\)

Vậy tập nghiệm của bất phương trình là: \(S = \left( { - \infty ; - \sqrt 2 } \right) \cup \left( {\sqrt 2 ; + \infty } \right).\)

Chọn D.