`2/(1-2x) in ZZ`
`-> 2 vdots 1 -2x`
`-> 1 - 2x in Ư(2)`
Vì `1 - 2x cancel vdots 2`
`-> 1 - 2x in {+-1}`
`-> x in {0, 1}`
Vì \(\dfrac{2}{1-2x}\in Z\Rightarrow2⋮1-2x\Rightarrow1-2x\in\left\{-2,-1,1,2\right\}\Rightarrow x\in\left\{1,0\right\}\) (vì \(x\in Z\))
Vậy x ∈ {1,0}