x \(\in\) Z \(\Leftrightarrow\) \(\frac{4-x}{x-1}\) \(\in\) Z
\(\Leftrightarrow\) \(\frac{4+\left(x-1\right)-x+1}{x-1}\) \(\in\) Z
\(\Leftrightarrow\) 1 + \(\frac{4-\left(x-1\right)}{x-1}\) \(\in\) Z
\(\Leftrightarrow\) \(\frac{4-\left(x-1\right)}{x-1}\) \(\in\) Z
\(\Leftrightarrow\) 1 + \(\frac{4}{x-1}\) \(\in\) Z
\(\Leftrightarrow\) \(\frac{4}{x-1}\) \(\in\) Z
\(\Leftrightarrow\) x - 1 \(\in\) Ư(4) = \(\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\) x \(\in\) \(\left\{2;0;3;-1;5;-3\right\}\)
Vậy x \(\in\) \(\left\{2;0;3;-1;5;-3\right\}\)
