Tập xác định: $\text{D}=ℝ\backslash \left\{2\right\}$. Đạo hàm: ${\text{y}}^{\text{'}}=1+\frac{\text{m}-1}{{(\text{x}-2)}^2}=\frac{{\text{x}}^2-4\text{x}+\text{m}+3}{{(\text{x}-2)}^2}$.

Xét hàm số $\text{f}\left(\text{x}\right)={\text{x}}^2-4\text{x}+3$ trên $[5;+\infty )$.

Có ${\text{f}}^{\text{'}}\left(\text{x}\right)=2\text{x}-4\Rightarrow$ trên khoảng $[5;+\infty )$ hàm số $\text{f}\left(\text{x}\right)={\text{x}}^2-4\text{x}+3$ đồng biến.

Khi đó ${\text{y}}^{\text{'}}\ge 0,\forall \text{x}\in [5;+\infty )\iff \text{f}\left(\text{x}\right)\ge -\text{m},\forall \text{x}\in [5;+\infty )$.

$\iff \text{f}\left(5\right)\ge -\text{m}\iff 8\ge -\text{m}\iff \text{m}\ge -8$.

Mà m nguyên âm nên $\text{m}\in \{-8;-7;-6;-5;-4;-3;-2;-1\}$.