Ta có:
$\dfrac{27^x}{3^{2x-y}} = \dfrac{3^{3x}}{3^{2x-y}} = 3^{3x-2x+y} = 3^{x+y} = 243 = 3^5$
$⇒ x+y=5$ ($1$)
$\dfrac{25^x}{5^{x+y}} = \dfrac{5^{2x}}{5^{x+y}} = 5^{2x-x-y} = 5^{x-y} = 125 = 5^3$
$⇒ x-y=3$ ($2$)
Từ ($1$);($2$) $⇒$ $\left\{\begin{matrix}x = (5+3):2 = 4 & \\ y = 5-4=1 & \end{matrix}\right.$
Vậy `(x;y)=(4;1)`