Đáp án:
$\begin{array}{l}
a)A = \dfrac{5}{3}\\
= > \dfrac{{n - 1}}{{n - 3}} = \dfrac{5}{3}\\
= > 5.\left( {n - 3} \right) = 3.\left( {n - 1} \right)\\
= > 5.n - 5.3 = 3.n - 3.1\\
= > 5.n - 15 = 3.n - 3\\
= > 5.n - 3.n = - 3 + 15\\
= > 2.n = 12\\
= > n = 6\left( {tmdk} \right)
\end{array}$
Vậy n=6
$\begin{array}{l}
b)A = \dfrac{{n - 1}}{{n - 3}}\\
= \dfrac{{n - 3 + 2}}{{n - 3}}\\
= 1 + \dfrac{2}{{n - 3}}\\
A\,đạt\,GTLN\\
= > \dfrac{2}{{n - 3}}\,đạt\,GTLN\\
= > n - 3\,đạt\,GTNN\\
Do:\dfrac{2}{{n - 3}} \in Z\\
= > n - 3\, \in \left\{ { - 2; - 1;1;2} \right\}\\
Khi:\,\left( {n - 3} \right)\,đạt\,GTNN\\
= > n - 3 = - 2\\
= > n = 1\left( {tm} \right)
\end{array}$
Vậy n=1