a) điều kiện \(ne-3\)

\(A=\dfrac{n-5}{n+3}=\dfrac{n+3-8}{n+3}=1-\dfrac{8}{n+3}\)

ta có : A nguyên \(\Leftrightarrow\dfrac{n-5}{n+3}\) nguyên \(\Leftrightarrow1-\dfrac{8}{n+3}\) nguyên \(\Leftrightarrow\dfrac{8}{n+3}\)

\(\Leftrightarrow n+3\) thuộc ước của 8 là \(\pm1;\pm2;\pm4;\pm8\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}n+3=1\+3=-1\end{matrix}\right.\\\left[{}\begin{matrix}n+3=2\+3=-2\end{matrix}\right.\\\left[{}\begin{matrix}n+3=4\+3=-4\end{matrix}\right.\\\left[{}\begin{matrix}n+3=8\+3=-8\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}n=-2\=-4\end{matrix}\right.\\\left[{}\begin{matrix}n=-1\=-5\end{matrix}\right.\\\left[{}\begin{matrix}n=1\=-7\end{matrix}\right.\\\left[{}\begin{matrix}n=5\=-11\end{matrix}\right.\end{matrix}\right.\) vậy ===-

b) ta có \(A< 0\Leftrightarrow\dfrac{n-5}{n+3}< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}n-5>0\+3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}n-5< 0\+3>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}n>5\< -3\end{matrix}\right.\\\left\{{}\begin{matrix}n< 5\>-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n\in\varnothing\\-3< n< 5\end{matrix}\right.\) vậy \(-3< n< 5\) thì \(A< 0\)