Đáp án:
$\begin{array}{l}
{x^2} - 2mx - 2m - 1 = 0\\
\Delta ' \ge 0\\
\Rightarrow {m^2} + 2m + 1 \ge 0\\
\Rightarrow {\left( {m + 1} \right)^2} \ge 0\left( {ld} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m\\
{x_1}{x_2} = - 2m - 1
\end{array} \right.\\
DO:{x_1} - 3{x_2} = 14\\
\Rightarrow {x_1} = 3{x_2} + 14\\
\Rightarrow 3{x_2} + 14 + {x_2} = 2m\\
\Rightarrow 4{x_2} = 2m - 14\\
\Rightarrow {x_2} = \dfrac{{m - 7}}{2}\\
\Rightarrow {x_1} = 3{x_2} + 14 = \dfrac{{3m - 21}}{2} + 14 = \dfrac{{3m + 7}}{2}\\
\Rightarrow \dfrac{{3m + 7}}{2}.\dfrac{{m - 7}}{2} = - 2m - 1\\
\Rightarrow 3{m^2} - 21m + 7m - 49 = - 8m - 4\\
\Rightarrow 3{m^2} - 6m - 45 = 0\\
\Rightarrow {m^2} - 2m - 15 = 0\\
\Rightarrow \left( {m - 5} \right)\left( {m + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 5\\
m = - 3
\end{array} \right.
\end{array}$
Vậy m=-3 hoặc m=5
