`text{đkxđ} x-2 \ne 0 => x \ne 2`
`text{ta có}`
`A=(6x-2)/(x-2) = (6(x-2)+10)/(x-2)= 6 + 10/(x-2)`
`text{để} A \in Z => 6 + 10/(x-2) \in Z => 10/(x-2) \in Z`
`=>10 \vdots x-2 => x-2 \in Ư_(10) = {+-1;+-2;+-5;+-10}`
`=>x \in {-8;0;1;+-3;4;7;12}`
`text{Vậy để A nguyên thì} x \in {-8;0;1;+-3;4;7;12}`
~~~~~~~~~~~GOOD LUCK~~~~~~~~~~~
