Đáp án:
a) f(x)=0 có hai nghiệm trái dấu thì:
$\begin{array}{l}
\left( {2m - 3} \right).\left( {m + 7} \right) < 0\\
\Rightarrow - 7 < m < \frac{3}{2}
\end{array}$
b) f(x)<0 vô nghiệm
$\begin{array}{l}
\Rightarrow f\left( x \right) \ge 0\forall x\\
\Rightarrow \left( {2m - 3} \right){x^2} + 2\left( {m + 1} \right)x + m + 7 \ge 0\forall x\\
\Rightarrow \left\{ \begin{array}{l}
2m - 3 > 0\\
\Delta ' \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > \frac{3}{2}\\
{\left( {m + 1} \right)^2} - \left( {2m - 3} \right)\left( {m + 7} \right) \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > \frac{3}{2}\\
{m^2} + 2m + 1 - 2{m^2} - 11m + 21 \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > \frac{3}{2}\\
- {m^2} - 9m + 22 \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > \frac{3}{2}\\
\left[ \begin{array}{l}
m > 2\\
m < - 11
\end{array} \right.
\end{array} \right.\\
\Rightarrow m > 2
\end{array}$
Vậy m>2
