Ans: D
$\begin{array}{cl}
& \dfrac{1}{n+3} – \dfrac{1}{3-n} \\
= & \dfrac{1}{3+n} – \dfrac{1}{3-n} \\
= & \dfrac{(3-n)-(3+n)}{(3-n)(3+n)} \\
= & \dfrac{3-n-3-n}{9-n^2} \\
= & \dfrac{-2n}{9-n^2} \\
= & \dfrac{2n}{n^2-9}
\end{array}$
$\begin{array}{cl}
& \dfrac{1}{n+3} – \dfrac{1}{3-n} \\
= & \dfrac{1}{3+n} – \dfrac{1}{3-n} \\
= & \dfrac{(3-n)-(3+n)}{(3-n)(3+n)} \\
= & \dfrac{3-n-3-n}{9-n^2} \\
= & \dfrac{-2n}{9-n^2} \\
= & \dfrac{2n}{n^2-9}
\end{array}$
