Ta có: \(2x +1 > 1 – x ⇔ x>0\). Vậy (A) sai
Xét (B), dễ thấy \((-4 + 1) (1 + 2) < 0 < (-2)^2\).
Vậy (B) đúng
+) Xét đáp án C:
\(\begin{array}{l}
\frac{1}{{1 - x}} + 2 \le 0 \Leftrightarrow \frac{{1 + 2 - 2x}}{{1 - x}} \le 0\\
\Leftrightarrow \frac{{3 - 2x}}{{1 - x}} \le 0 \Leftrightarrow \frac{{2x - 3}}{{x - 1}} \le 0\\
\Leftrightarrow 1 < x \le \frac{3}{2} \Rightarrow x \in \left( {1;\;\frac{3}{2}} \right)\\ \Rightarrow x = - 2 \notin \left( {1;\;\frac{3}{2}} \right).
\end{array}\)
\(\Rightarrow \) Đáp án C sai.
+) Xét đáp án D:
\(\begin{array}{l}
\left( {2 - x} \right){\left( {x + 2} \right)^2} < 0\\
\Leftrightarrow \left\{ \begin{array}{l}
2 - x < 0\\
x + 2 \ne 0
\end{array} \right.\;\;\;\left( {do\;\;{{\left( {x + 2} \right)}^2} \ge 0} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
x > 2\\
x \ne - 2
\end{array} \right. \Leftrightarrow x > 2 \\\Rightarrow x = - 2 \notin \left( {2; + \infty } \right).
\end{array}\)
\(\Rightarrow \) Đáp án D sai.
