A = \(\frac{n+2}{n-5}=\frac{\left(n-5\right)+7}{n-5}=1+\frac{7}{n-5}\)
A \(\in\)Z => \(\frac{7}{n-5}\in\)Z => 7 \(⋮\)n - 5 => n - 5 \(\in\)Ư(7) = {\(\pm\)1; \(\pm\)7}
Vậy n \(\in\){-2; 4; 6; 12} thì A \(\in\)Z
Đề là A\(\in\)Z ko phải A = Z bạn nhé!
\(A\in Z\Rightarrow n+2⋮n-5\)
\(\Rightarrow\left(n-5\right)+7⋮n-5\)
\(\Rightarrow7⋮n+5\)
\(\Rightarrow n-5\in\left\{1;-1;7;-7\right\}\)
\(\Rightarrow n\in\left\{6;4;12;-2\right\}\)