Đáp án: $\,m = \dfrac{{ - 2 \pm \sqrt {31} }}{6}$
Giải thích các bước giải:
$\begin{array}{l}
{x^2} - \left( {2m - 1} \right).x - 2m = 0\\
\Delta > 0\\
\Leftrightarrow {\left( {2m - 1} \right)^2} - 4.\left( { - 2m} \right) > 0\\
\Leftrightarrow 4{m^2} - 4m + 1 + 8m > 0\\
\Leftrightarrow {\left( {2m + 1} \right)^2} > 0\\
\Leftrightarrow m\# - \dfrac{1}{2}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 1\\
{x_1}{x_2} = - 2m
\end{array} \right.\\
x_1^2 + x_2^2 - 4{x_1}{x_2} + 2{x_1}\left( {{x_1} - {x_2}} \right) + 2{x_2}\left( {{x_1} + {x_2}} \right) = 12\\
\Leftrightarrow 3x_1^2 + 3x_2^2 - 4{x_1}{x_2} - 12 = 0\\
\Leftrightarrow 3\left( {x_1^2 + x_2^2 + 2{x_1}{x_2}} \right) - 6{x_1}{x_2} - 4{x_1}{x_2} - 12 = 0\\
\Leftrightarrow 3{\left( {{x_1} + {x_2}} \right)^2} - 10{x_1}{x_2} - 12 = 0\\
\Leftrightarrow 3.{\left( {2m - 1} \right)^2} - 10.\left( { - 2m} \right) - 12 = 0\\
\Leftrightarrow 12{m^2} - 12m + 3 + 20m - 12 = 0\\
\Leftrightarrow 12{m^2} + 8m - 9 = 0\\
\Leftrightarrow m = \dfrac{{ - 2 \pm \sqrt {31} }}{6}\left( {tmdk} \right)\\
Vay\,m = \dfrac{{ - 2 \pm \sqrt {31} }}{6}
\end{array}$
