Đáp án:
$\begin{array}{l}
{x^2} + mx - 3 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {m^2} - 4.\left( { - 3} \right) > 0\\
\Rightarrow {m^2} + 12 > 0\left( {ld} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - m\\
{x_1}{x_2} = - 3
\end{array} \right.\\
\left| {{x_1}} \right| + \left| {{x_2}} \right| = 4\\
\Rightarrow x_1^2 + 2\left| {{x_1}{x_2}} \right| + x_2^2 = 16\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} + 2.\left| { - 3} \right| = 16\\
\Rightarrow {m^2} - 2.\left( { - 3} \right) + 6 = 16\\
\Rightarrow {m^2} = 16\\
\Rightarrow m = \pm 4
\end{array}$
