Đáp án:
$\begin{array}{l}
a)m{x^2} - 2\left( {m - 1} \right)x + m + 2 = 0\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
\Delta ' > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
{\left( {m - 1} \right)^2} - m\left( {m + 2} \right) > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
{m^2} - 2m + 1 - {m^2} - 2m > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
m < \dfrac{1}{4}
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_1}{x_2} = m + 2
\end{array} \right.\\
{x_1} + {x_2} = - 4{x_1}{x_2}\\
\Rightarrow 2\left( {m - 1} \right) = - 4\left( {m + 2} \right)\\
\Rightarrow m - 1 = - 2m - 4\\
\Rightarrow 3m = - 3\\
\Rightarrow m = - 1\left( {tmdk} \right)\\
\text{Vậy}\,m = - 1\\
b)\\
+ Khi:m = 0\\
\Rightarrow 2x + 2 = 0\\
\Rightarrow x = - 1\left( {tm} \right)\\
+ Khi:m \ne 0\\
\Rightarrow \Delta ' \ge 0\\
\Rightarrow - 4m + 1 \ge 0\\
\Rightarrow m \le \dfrac{1}{4}\\
\text{Vậy}\,m \le \dfrac{1}{4}\,thi\,pt\,\text{có}\,\text{nghiệm}
\end{array}$
