Đáp án: $m = - 3$
Giải thích các bước giải:
$\begin{array}{l}
{x^2} - \left( {2m + 1} \right)x - 3 = 0\\
\Leftrightarrow \Delta = {\left( {2m + 1} \right)^2} - 4.\left( { - 3} \right)\\
= {\left( {2m + 1} \right)^2} + 12 > 0
\end{array}$
Nên pt luôn có 2 nghiệm phân biệt với mọi m
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 1\\
{x_1}{x_2} = - 3
\end{array} \right.\\
Khi:\left\{ \begin{array}{l}
{x_1} < {x_2}\\
\left| {{x_1}} \right| - \left| {{x_2}} \right| = 5 > 0
\end{array} \right.\\
\Leftrightarrow {x_1} < 0 < {x_2}\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} < 0\\
{x_1}.{x_2} < 0\\
x_1^2 - 2\left| {{x_1}{x_2}} \right| + x_2^2 = 25
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2m + 1 < 0\\
- 3 < 0\left( {tm} \right)\\
{\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} - 2.\left| {{x_1}{x_2}} \right| = 25
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < \frac{{ - 1}}{2}\\
{\left( {2m + 1} \right)^2} - 2.\left( { - 3} \right) - 2.\left| { - 3} \right| = 25
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < - \frac{1}{2}\\
{\left( {2m + 1} \right)^2} = 25
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < \frac{{ - 1}}{2}\\
\left[ \begin{array}{l}
2m + 1 = 5\\
2m + 1 = - 5
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < - \frac{1}{2}\\
\left[ \begin{array}{l}
m = 2\left( {ktm} \right)\\
m = - 3\left( {tm} \right)
\end{array} \right.
\end{array} \right.\\
Vậy\,m = - 3
\end{array}$
