Đặt $\dfrac{x}{3}=\dfrac{y}{5}=k$
$\to \begin{cases}x=3k\\y=5k\end{cases}$
Thế $x=3k;y=5k$ vào $A$:
$A=\dfrac{5.(3k)^2+3(5k)^2}{10.(3k)^2-3(5k)^2}$
$=\dfrac{5.9k^2+3.25k^2}{10.9k^2-3.25k^2}$
$=\dfrac{45k^2+75k^2}{90k^2-75k^2}$
$=\dfrac{120k^2}{15k^2}=8$
Vậy $A=8$ khi $\dfrac{x}{3}=\dfrac{y}{5}$