Phương pháp giải:

\(\left| {f\left( x \right)} \right| \le g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l}g\left( x \right) \ge 0\\ - g\left( x \right) \le f\left( x \right) \le g\left( x \right)\end{array} \right.\)

Lời giải chi tiết:

\(\begin{array}{l}\left| {2x + 5} \right| \le {x^2} + 2x + 4 \Leftrightarrow \left\{ \begin{array}{l} - {x^2} - 2x - 4 \le 2x + 5\\2x + 5 \le {x^2} + 2x + 4\end{array} \right.\,\,\,\,\left( {{x^2} + 2x + 4 = {{\left( {x + 1} \right)}^2} + 3 > 0\forall x} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}{x^2} + 4x + 9 \ge 0\\{x^2} - 1 \ge 0\end{array} \right.\,\,\,\left( {do\,\,\,\,{x^2} + 4x + 9 = {{\left( {x + 2} \right)}^2} + 5 > 0\forall x} \right)\\ \Leftrightarrow {x^2} - 1 \ge 0 \Leftrightarrow {x^2} \ge 1 \Leftrightarrow \left[ \begin{array}{l}x \ge 1\\x \le - 1\end{array} \right.\end{array}\)

Chọn A.