$x_{1}^2-5x_{1}x_{2}+x_{2}^2=25$
$⇔(x_{1}+x_{2})^2-7x_{1}x_{2}-25=0$
Theo định lí Vi-ét, ta có:
$x_{1}+x_{2}=2m$
$x_{1}x_{2}=2m-1$
$⇒(x_{1}+x_{2})^2-7x_{1}x_{2}-25=0$
$⇔4m^2-7(2m-1)-25=0$
$⇔4m^2-14m-18=0$
$⇔\left[ \begin{array}{l}m=\frac{9}{2}\\m=-1\end{array} \right.$.