\(M=\dfrac{x^2-3m+5}{m-1}=\dfrac{m^2-m-2m+2+3}{m-1}=\dfrac{m\left(m-1\right)-2\left(m-1\right)+3}{m-1}=\dfrac{\left(m-2\right)\left(m-1\right)+3}{m-1}=m-2+\dfrac{3}{m-1}\)Để M nguyên
⇒ \(\dfrac{3}{m-1}\in Z\Rightarrow3⋮\left(m-1\right)\)
\(\Rightarrow m-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
⇒ \(m\in\left\{2;0;4;-2\right\}\)
Mà m là số tự nhiên
⇒ \(m\in\left\{0;2;4\right\}\)