Đáp án:
a) Để pt có 2 nghiệm dương thì:
$\begin{array}{l}
\left\{ \begin{array}{l}
\Delta ' > 0\\
\frac{{ - b}}{a} > 0\\
\frac{c}{a} > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\left( {m + 3} \right)^2} - 4m + 1 > 0\\
2\left( {m + 3} \right) > 0\\
4m - 1 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{m^2} + 6m + 9 - 4m + 1 > 0\\
m > - 3\\
m > \frac{1}{4}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{m^2} + 2m + 10 > 0\left( {luon\,dung} \right)\\
m > \frac{1}{4}
\end{array} \right.\\
Vậy\,m > \frac{1}{4}
\end{array}$
b) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 3} \right)\\
{x_1}{x_2} = 4m - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2{x_1} + 2{x_2} = 4m + 12\\
{x_1}{x_2} = 4m - 1
\end{array} \right.\\
\Rightarrow 2{x_1} + 2{x_2} - {x_1}{x_2} = 13
\end{array}$
