$25m^2-2x<m^2x-25\Leftrightarrow x(m^2+2)>25m^2+25$$\Leftrightarrow x>\frac{25m^2+25}{m^2+2}$
ta có: $x\in (15;+\infty) $ nên $\frac{25m^2+25}{m^2+2}\geq 15$
$\Leftrightarrow 10m^2\geq 5\Leftrightarrow m\geq \frac{1}{\sqrt2}$ hoặc $m\leq -\frac{1}{\sqrt2}$
Vậy $m\in(-\infty ;-\frac{1}{\sqrt2}]\cup [\frac{1}{\sqrt2};+\infty )$