Đáp án:
$\begin{array}{l}
a)\left( {m - 2} \right).{x^2} + 2\left( {m - 2} \right).x + 1 = 0\\
Cho:m - 2 = 0 \Rightarrow m = 2\\
\Rightarrow 0.{x^2} + 2.0.x + 1 = 0\\
\Rightarrow 1 = 0\left( {vn} \right)\\
+ Cho:m - 2 \ne 0 \Rightarrow m \ne 2\\
\Rightarrow \Delta ' < 0\\
\Rightarrow {\left( {m - 2} \right)^2} - \left( {m - 2} \right) < 0\\
\Rightarrow \left( {m - 2} \right)\left( {m - 2 - 1} \right) < 0\\
\Rightarrow \left( {m - 2} \right)\left( {m - 3} \right) < 0\\
\Rightarrow 2 < m < 3\\
Vay\,2 \le m < 3\\
b)Khi:m = 1\\
\Rightarrow 0.{x^2} - 6.0.x + 2 - 3 = 0\\
\Rightarrow - 1 = 0\left( {ktm} \right)\\
Khi:m \ne 1\\
\Rightarrow \Delta ' > 0\\
\Rightarrow 9{\left( {m - 1} \right)^2} - \left( {m - 1} \right)\left( {2m - 3} \right) > 0\\
\Rightarrow \left( {m - 1} \right)\left( {9m - 9 - 2m + 3} \right) > 0\\
\Rightarrow \left( {m - 1} \right).\left( {7m - 6} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
m > 1\\
m < \dfrac{6}{7}
\end{array} \right.\\
Vay\,m > 1\,hoac\,m < \dfrac{6}{7}\\
c)a.c < 0\\
\Rightarrow m.\left( {m - 5} \right) < 0\\
\Rightarrow 0 < m < 5\\
d)\left\{ \begin{array}{l}
m \ne 3\\
\Delta ' > 0\\
\dfrac{{ - b}}{a} > 0\\
\dfrac{c}{a} > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
m \ne 3\\
{\left( {m + 3} \right)^2} - \left( {3 - m} \right)\left( {m + 2} \right) > 0\\
\dfrac{{2\left( {m + 3} \right)}}{{3 - m}} > 0\\
\dfrac{{m + 2}}{{3 - m}} > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 3\\
{m^2} + 6m + 9 + {m^2} - m - 6 > 0\\
- 3 < m < 3\\
- 2 < m < 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 3\\
2{m^2} + 5m + 3 > 0\\
- 2 < m < 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 3\\
\left( {2m + 3} \right)\left( {m + 1} \right) > 0\\
- 2 < m < 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 2 < m < 3\\
\left[ \begin{array}{l}
m > - 1\\
m < - \dfrac{3}{2}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
- 2 < m < - \dfrac{3}{2}\\
- 1 < m < 3
\end{array} \right.
\end{array}$
