De obicei astfel de exercitii se fac in doua moduri:
Metoda 1:
[tex]\text{Face un artificiu de calcul:}\\
\dfrac{2x+1}{x-1}=\dfrac{2x-2+3}{x-1}=\dfrac{2(x-1)+3}{x-1}=\dfrac{2(x-1)}{x-1}+\dfrac{3}{x-1}=2+\dfrac{3}{x-1}\\
\text{Deoarece }2\in \mathbb{Z}\Rightarrow \dfrac{3}{x-1}\in \mathbb{Z}\\
\text{Asta inseamna ca x-1 trebuie sa fie divizor a lui 3:}\\
x-1\in D_3\\
x-1 \in \{3,1,-1,-3\}\\
x \in \{4,2,0,-2\}[/tex]
Metoda 2:
[tex]\text{Daca fractia este un numar intreg,inseamna ca numitorul se divide }\\
\text{ cu numaratorul,altfel spus :}\\
x-1\ |\ x-1\Rightarrow x-1\ |\ 2x-2\\
x-1\ |\ 2x+1\Rightarrow x-1\ |\ 2x+1\\
\text{Scazand relatiile se obtine:}\\
x-1\ | -3\Rightarrow x-1 \in D_{-3}\\
\text{Si am ajuns exact ca la prima metoda.}[/tex]