$y = 2x^3 - 3(2m+1)x^2 + (6m^2+6m)x +1$
$y' = 6x^2 - 6(2m+1)x + 6m^2 + 6m$
Hàm số đồng biến trên $(2;+\infty)$
$\to y' \geq 0\quad \forall x\in(2;+\infty)$
$\to x^2 - (2m +1)x + m^2+ m \geq 0\quad \forall x\in(2;+\infty)$
$\to (x-m)(x - m - 1)\geq 0\quad \forall x\in(2;+\infty)$
$\to (2-m)(2- m - 1) \geq 0$
$\to (2-m)(1-m)\geq 0$
$\to \left[\begin{array}{l}m\geq 2\\m \leq 1\end{array}\right.$