$\dfrac{x^2-x+2}{x-1}=$ $\dfrac{x(x-1)+2}{x-1}$ $=\dfrac{x(x-1)}{x-1}+$ $\dfrac{2}{x-1}$ $=x+\dfrac{2}{x-1}$
$\text{⇒ 2 $\vdots$ x-1 ⇒ x-1 ∈ $Ư_{(2)}$={±1;±2}}$
$\left[\begin{array}{ccc}x-1&-1&1&-2&2\\x&0&2&-1&3\end{array}\right]$
phấn b làm tương tự