Đáp án:
a.\(m \in \left[ { - 3;9} \right]\)
b. \(m \in \left( { - \infty ; - \frac{1}{3}} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a.{x^2} + (m + 1)x + 2m + 7 \ge 0\\
\Leftrightarrow \left\{ \begin{array}{l}
1 > 0\left( {ld} \right)\\
{m^2} + 2m + 1 - 8m - 28 \le 0
\end{array} \right.\\
\to {m^2} - 6m - 27 \le 0\\
\to m \in \left[ { - 3;9} \right]\\
b.m{x^2} + (m - 1)x + m - 1 < 0\\
\to \left\{ \begin{array}{l}
m < 0\\
{m^2} - 2m + 1 - 4m\left( {m - 1} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 0\\
{m^2} - 2m + 1 - 4{m^2} + 4m < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 0\\
- 3{m^2} + 2m + 1 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 0\\
m \in \left( { - \infty ; - \frac{1}{3}} \right) \cup \left( {1; + \infty } \right)
\end{array} \right.\\
KL:m \in \left( { - \infty ; - \frac{1}{3}} \right)
\end{array}\)
