Để D= \(\frac{2n+7}{n+3}\) \(\in\) Z
thì 2n+7 \(⋮\) n+3
\(\Leftrightarrow\) (2n + 6)+ 1 \(⋮\) n+ 3
\(\Leftrightarrow\) 2 ( n+ 3 ) + 1 \(⋮\) n+3
\(\Leftrightarrow\) 1 \(⋮\) n+3
\(\Leftrightarrow\) n+ 3 \(\in\) Ư (1)
\(\Leftrightarrow\) n+3= -1; 1
\(\Leftrightarrow\) n = -4 ; -2