Đáp án: m>-1
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
x + 4{m^2} \le 2mx + 1\\
3x + 2 > 2x + 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {2m - 1} \right)x \ge 4{m^2} - 1\\
x > - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {2m - 1} \right)x \ge \left( {2m - 1} \right)\left( {2m + 1} \right)\\
x > - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2m + 1;2m > 1\\
x \le 2m + 1;2m < 1\\
2m - 1 = 0
\end{array} \right.\\
x > - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \ge 2m + 1;x > - 1;m > \frac{1}{2}\\
x \le 2m + 1;x > - 1;m < \frac{1}{2}\\
m = \frac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m \ge \frac{1}{2}\\
2m + 1 > - 1;m < \frac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m \ge \frac{1}{2}\\
m > - 1
\end{array} \right.\\
\Rightarrow m > - 1
\end{array}$
