[tex]A=\Big\{x\in\mathbb_{Z}$ $\Big|$ $ \dfrac{3x-5}{x-1} \in \mathbb_{Z}$ $\Big\}\\ \\ \dfrac{3x-5}{x-1} = \dfrac{3x-3-2}{x-1} = \dfrac{3(x-1)-2}{x-1} = \dfrac{3(x-1)}{x-1}-\dfrac{2}{x-1} = \underset{\in \mathbb_{Z}}{3}-\dfrac{2}{x-1}$ $ \\ \\ \\ \bullet \dfrac{2}{x-1}\in \mathbb_{Z} \Rightarrow $ $x-1\in D_2\Rightarrow x-1\in \Big\{\pm1,\pm2\Big\}\Rightarrow \\ \\ \Rightarrow x-1 \in \Big\{-2,-1,1,2\Big\} \Big|+1 \Rightarrow x\in \Big\{-1,0,2,3\Big\} \Rightarrow[/tex]
[tex] \Rightarrow \boxed{A = \Big\{-1,0,2,3\Big\}}[/tex]
