Đáp án:
$\begin{array}{l}
1b)\\
\left| {x + 1} \right| - \left| {x - 2} \right| \ge 3\\
+ Khi:x \ge 2\\
\Rightarrow x + 1 - \left( {x - 2} \right) \ge 3\\
\Rightarrow 3 \ge 3\left( {luon\,dung} \right)\\
+ Khi: - 1 \le x < 2\\
\Rightarrow x + 1 - \left( {2 - x} \right) \ge 3\\
\Rightarrow x + 1 - 2 + 2x \ge 3\\
\Rightarrow 3x \ge 4\\
\Rightarrow x \ge \frac{4}{3}\\
+ Khi:x < - 1\\
\Rightarrow - x - 1 - \left( {2 - x} \right) \ge 3\\
\Rightarrow - x - 1 - 2 + 2x \ge 3\\
\Rightarrow x \ge 6\left( {ktm} \right)\\
Vậy\,x \ge \frac{4}{3}\\
2)\\
\left( {m + 4} \right){x^2} - \left( {m - 4} \right)x - 2m + 1 \ge 0\forall x\\
\Rightarrow \left\{ \begin{array}{l}
m + 4 > 0\\
\Delta \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > - 4\\
{\left( {m - 4} \right)^2} - 4\left( {m + 4} \right)\left( { - 2m + 1} \right) \le 0
\end{array} \right.\\
\Rightarrow {m^2} - 8m + 16 + 8{m^2} + 28m - 16 \le 0\\
\Rightarrow 9{m^2} + 20m \le 0\\
\Rightarrow - \frac{{20}}{9} \le m \le 0
\end{array}$
