Đáp án:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
2 - m \ne 0\\
m + 1 \ne 0
\end{array} \right. \Rightarrow m \ne 2;m \ne - 1\\
\left( 1 \right)\left( {2 - m} \right)x + 2m + 4 \ge 0\\
\Rightarrow \left( {2 - m} \right)x \ge - 2m - 4\\
\Rightarrow \left[ \begin{array}{l}
x \ge \frac{{ - 2m - 4}}{{2 - m}}\left( {khi:2 - m > 0 \Rightarrow m < 2} \right)\\
x \le \frac{{ - 2m - 4}}{{2 - m}}\left( {khi:m > 2} \right)
\end{array} \right.\\
\left( 2 \right):\left( {m + 1} \right)x + {m^2} - 4 \ge 0\\
\Rightarrow \left[ \begin{array}{l}
x \ge \frac{{4 - {m^2}}}{{m + 1}}\left( {khi:m > - 1} \right)\\
x \le \frac{{4 - {m^2}}}{{m + 1}}\left( {khi:m < - 1} \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( 2 \right)\\
\Rightarrow \frac{{ - 2m - 4}}{{2 - m}} = \frac{{4 - {m^2}}}{{m + 1}}\left( {khi: - 1 < m < 2} \right)\\
\Rightarrow \frac{{ - 2}}{{2 - m}} = \frac{{2 - m}}{{m + 1}}\\
\Rightarrow {m^2} - 4m + 4 = - 2m - 2\\
\Rightarrow {m^2} - 2m + 6 = 0\\
\Rightarrow m \in \emptyset
\end{array}$
Vậy ko có m thỏa mãn
