Đáp án:
$\begin{array}{l}
a)TXD:R/{\rm{\{ }}2\} \\
\Rightarrow \frac{{2x + m}}{{x + 2}} \ge 0\forall x \ne 2\\
\Rightarrow \left( {2x + m} \right) \vdots \left( {x + 2} \right)\\
\Rightarrow 2\left( {x + \frac{m}{2}} \right) \vdots \left( {x + 2} \right)\\
\Rightarrow \frac{m}{2} = 2\\
\Rightarrow m = 4\\
b)\left( {2x + 3} \right)\left( {x + m} \right) \ge 0\forall x\\
\Rightarrow \left( {x + \frac{3}{2}} \right)\left( {x + m} \right) \ge 0\forall x\\
\Rightarrow {x^2} + \left( {\frac{3}{2} + m} \right)x + \frac{3}{2}m \ge 0\forall x\\
\Rightarrow \Delta \le 0\\
\Rightarrow {\left( {m + \frac{3}{2}} \right)^2} - 4.\frac{3}{2} \le 0\\
\Rightarrow {m^2} + 3m + \frac{9}{4} - 6 \le 0\\
\Rightarrow {m^2} + 3m - \frac{{15}}{4} \le 0\\
\Rightarrow \frac{{ - 3 - \sqrt {69} }}{4} \le m \le \frac{{ - 3 + \sqrt {69} }}{4}
\end{array}$
