Để A là số nguyên thì: n2+4 \(⋮\) n+3
\(\Leftrightarrow n.n+4⋮n+3\)
\(\Leftrightarrow n.n+3n-3n-9+13⋮n+3\)
\(\Leftrightarrow n\left(n+3\right)-3\left(n+3\right)+13⋮n+3\)
\(\Leftrightarrow\left(n-3\right)\left(n+3\right)+13⋮n+3\)
Vì: \(\left(n-3\right)\left(n+3\right)⋮n+3\)
nên \(13⋮n+3\) \(\Rightarrow n+3\in\left\{1;-1;13;-13\right\}\)
\(\Rightarrow n\in\left\{-2;-4;10;-16\right\}\)
Vậy:.................
ĐKXĐ: \(n\ne-3\)
\(\frac{n^2+4}{n+3}=n-3+\frac{13}{n+3}\)
Để A nguyên thì \(\left(n+3\right)⋮13\Leftrightarrow\left(n+3\right)\inƯ\left(13\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}n+3=1\\n+3=-1\\n+3=13\\n+3=-13\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=-2\\n=-4\\n=10\\n=-16\end{matrix}\right.\)
Vậy ...