để A nguyên thì 4x-4 ⋮ x-2
vì (x-2) ⋮(x-2)
=> 4(x-2)⋮(x-2)
=> 4x-8⋮ (x-2)
=> (4x-8)-(4x-4) ⋮ (x-2)
=> (4x-8-4x+4) ⋮ (x-2)
=> -4⋮(x-2)
=> x-2 ∈Ư(-4)={-4;-2;-1;2;4}
=> x ∈ {-2;0;1;4;6}
vậy x ∈ {-2;0;1;4;6}thì A nguyên
Để \(A=\dfrac{4x-4}{x-2}\) có giá trị nguyên thì \(4x-4⋮x-2\\ \Rightarrow4x-4-4\left(x-2\right)⋮x-2\\ \Rightarrow4x-4-4x+8⋮x-2\\ \Rightarrow4⋮x-2\\ \Rightarrow x-2\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow x\in\left\{-2;0;1;3;4;6\right\}\)
Vậy với \(x\in\left\{-2;0;1;3;4;6\right\}\) thì \(A=\dfrac{4x-4}{x-2}\) có giá trị nguyên.
