$A=\dfrac{1-2x}{x+3}=\dfrac{-2x-6x+7}{x+3}=\dfrac{-2(x+3)+7}{x+3} \\ A=\dfrac{-2(x+3)}{x+3}+\dfrac{7}{x+3} = -2+\dfrac{7}{x+3}$
Để $A \in \mathbb{Z} $ thì:
$x+3 \in U(7) =\left \{ \pm 1;\pm7 \right \} \\ x+3=-1 \to x=-4 \\ x+3=1 \to x=-2 \\ x+3=7 \to x=4 \\ x+3=-7 \to x=-10$