Đáp án:
$\begin{array}{l}
1)m{x^2} - 2\left( {m + 2} \right).x + 2 + 3m = 0\\
+ Khi:m = 0\\
\Rightarrow - 4x + 2 = 0\\
\Rightarrow x = \frac{1}{2}\left( {ktm} \right)\\
+ Khi:m \ne 0\\
\Rightarrow \Delta ' < 0\\
\Rightarrow {\left( {m + 2} \right)^2} - m.\left( {2 + 3m} \right) < 0\\
\Rightarrow {m^2} + 4m + 4 - 2m - 3{m^2} < 0\\
\Rightarrow 2{m^2} - 2m - 4 > 0\\
\Rightarrow {m^2} - m - 2 > 0\\
\Rightarrow \left( {m - 2} \right)\left( {m + 1} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
m > 2\\
m < - 1
\end{array} \right.\\
Vay\,m > 2/m < - 1\\
2)\\
y \le 0\forall x\\
\Rightarrow m{x^2} - 2\left( {m + 3} \right)x + 3m - 1 \le 0\forall x\\
\Rightarrow \left\{ \begin{array}{l}
m < 0\\
\Delta ' \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m < 0\\
{\left( {m + 3} \right)^2} - m.\left( {3m - 1} \right) \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m < 0\\
- 2{m^2} + 7m + 9 \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m < 0\\
\left( {2m - 9} \right)\left( {m + 1} \right) \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m < 0\\
\left[ \begin{array}{l}
m \ge \frac{9}{2}\\
m \le - 1
\end{array} \right.
\end{array} \right.\\
\Rightarrow m \le - 1
\end{array}$
