Đáp án:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
m + 2 \ne 0\\
\Delta > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 2\\
{\left( {m + 4} \right)^2} - 4\left( {m + 2} \right)\left( {1 - m} \right) > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 2\\
{m^2} + 8m + 16 + 4{m^2} + 4m - 8 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 2\\
5{m^2} + 12m + 8 > 0\forall x
\end{array} \right.\\
\Rightarrow m \ne - 2\\
b)Theo\,viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{m + 4}}{{m + 2}} = 1 + \frac{2}{{m + 2}}\\
{x_1}{x_2} = \frac{{1 - m}}{{m + 2}} = \frac{{3 - 2 - m}}{{m + 2}} = \frac{3}{{m + 2}} - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
3\left( {{x_1} + {x_2}} \right) = 3 + \frac{6}{{m + 2}}\\
2{x_1}{x_2} = \frac{6}{{m + 2}} - 2
\end{array} \right.\\
\Rightarrow 3\left( {{x_1} + {x_2}} \right) - 2{x_1}{x_2} = 3 + \frac{6}{{m + 2}} - \frac{6}{{m + 2}} + 2 = 5\\
Vậy\,3\left( {{x_1} + {x_2}} \right) - 2{x_1}{x_2} = 5\,ko\,phụ\,thuộc\,vào\,m
\end{array}$
