[tex]\displaystyle\bf\\ \frac{2x+7}{x+1}=\frac{2x+2+5}{x+1}=\frac{2x+2}{x+1}+\frac{5}{x+1}=\frac{2(x+1)}{x+1}+\frac{5}{x+1}=2+\frac{5}{x+1}\\\\2\in N\implies~Ne~ocupam~de~~\frac{5}{x+1}\\\\\\\frac{5}{x+1}\in N~\implies~5~\vdots~(x+1)\\\\D_5=\{1;~5\}\\\\x+1=1\implies \boxed{\bf~x_1=0}\\\\x+1=5\implies \boxed{\bf~x_2=4}\\\\\\\boxed{\boxed{\bf~G=\{0;~4\}}}[/tex]
