Đáp án:
Dk để có 2 nghiệm pb là:
$\begin{array}{l}
\Delta ' > 0\\
\Rightarrow {\left( {m + 2} \right)^2} - {m^2} + 9 > 0\\
\Rightarrow {m^2} + 4m + 4 - {m^2} + 9 > 0\\
\Rightarrow 4m > - 13\\
\Rightarrow m > - \frac{{13}}{4}
\end{array}$
Theo Viet:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 4\\
{x_1}{x_2} = {m^2} - 9
\end{array} \right.\\
\left| {{x_1} - {x_2}} \right| = {x_1} + {x_2}\left( {dk:{x_1} + {x_2} > 0 \Rightarrow m > - 2} \right)\\
\Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = {\left( {{x_1} + {x_2}} \right)^2}\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = {\left( {{x_1} + {x_2}} \right)^2}\\
\Rightarrow {x_1}{x_2} = 0\\
\Rightarrow {m^2} - 9 = 0\\
\Rightarrow \left[ \begin{array}{l}
m = - 3\left( {ktm} \right)\\
m = 3\left( {tm} \right)
\end{array} \right.
\end{array}$
Vậy m=3