Đáp án:
b) \(m \in \emptyset \)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\left\{ \begin{array}{l}
m \ne - 2\\
{m^2} - 3\left( {m + 2} \right) \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 2\\
{m^2} - 3m - 6 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 2\\
\left[ \begin{array}{l}
m \ge \dfrac{{3 + \sqrt {33} }}{2}\\
m \le \dfrac{{3 - \sqrt {33} }}{2}
\end{array} \right.
\end{array} \right.\\
b)f\left( x \right) \le 0\\
\to \left( {m + 2} \right){x^2} + 2mx + 3 \le 0\\
\to \left\{ \begin{array}{l}
m + 2 < 0\\
{m^2} - 3\left( {m + 2} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < - 2\\
{m^2} - 3m - 6 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < - 2\\
m \in \left( {\dfrac{{3 - \sqrt {33} }}{2};\dfrac{{3 + \sqrt {33} }}{2}} \right)
\end{array} \right.\\
\to m \in \emptyset
\end{array}\)
