$A = \dfrac{2x^3 + 5x^2 - 5x + 5}{2x-1} \bigg(x \neq \dfrac{1}{2}\bigg)$
$= \dfrac{2x^3 - x^2 + 6x^2 - 3x - 2x + 1 +4}{2x-1}$
$=\dfrac{x^2(2x - 1) + 3x(2x -1) - (2x-1) + 4}{2x-1}$
$=\dfrac{(2x-1)(x^2 + 3x - 1) + 4}{2x - 1}$
Để A là số nguyên thì $\text{$(2x-1)(x^2 + 3x - 1) +4$ $\vdots$ 2x-1}$
⇔ $\text{4 $\vdots$ 2x-1}$
⇔ (2x-1) ∈ {1; -1; 2; -2; 4; -4}$
Tương ứng x ∈ {1; 0; 1,5; 2.5; -1.5}$
Mà x ∈ Z nên x ∈ {1; 0}$
