Đk : $x \ge 1 , y \ge 4$
$$M = \dfrac{y\sqrt{x-1} + x\sqrt{y-4}}{xy}=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-4}}{y}$$
AD Cosi : $$(x-1)+1 \ge 2\sqrt{x-1} \Longrightarrow \dfrac{\sqrt{x-1}}{x}\le \dfrac{1}{2}$$
$$(y-4)+4 \ge 4\sqrt{y-4}\Longrightarrow \dfrac{\sqrt{y-4}}{y} \le \dfrac{1}{4}$$
$$\Longrightarrow M \le \dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}$$
Dấu = tại $x=2 ,y=8$