Notez [tex]n=\dfrac{3x+5}{x-2}.[/tex]
[tex]n=\dfrac{3(x-2)+11}{x-2}=3+\dfrac{11}{x-2} \\ \\ \Rightarrow (x-2)\in\{\pm 1;\pm 11\}[/tex]
Luăm toate cazurile:
[tex]x-2=1\Rightarrow x=3 \Rightarrow n=14.[/tex]
[tex]x-2=-1\Rightarrow x=1\Rightarrow n=-8\text{ (nu este natural)}[/tex]
[tex]x-2=11\Rightarrow x=13 \Rightarrow n=4.[/tex]
[tex]x-2=-11\Rightarrow x=-9\Rightarrow n=2.[/tex]
În concluzie, [tex]x\in \{3,13,-9\}[/tex]