Đáp án:
$\begin{array}{l}
{x^2} - mx - 2 = 0\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = - 2
\end{array} \right.\\
\left| {{x_1} - {x_2}} \right| = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2}} \\
= \sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \\
= \sqrt {{m^2} - 4.\left( { - 2} \right)} \\
= \sqrt {{m^2} + 8} \\
Do:{m^2} + 8 \ge 8\forall m\\
\Rightarrow \sqrt {{m^2} + 8} \ge \sqrt 8 = 2\sqrt 2 \\
\Rightarrow \left| {{x_1} - {x_2}} \right| \ge 2\sqrt 2 \forall m
\end{array}$
