\(ĐKXĐ:x\ge3;y\ge2\)
\(A=\dfrac{x\sqrt{y-2}+y\sqrt{x-3}}{xy}=\dfrac{\dfrac{1}{\sqrt{2}}.x\sqrt{2\left(y-2\right)}+\dfrac{1}{\sqrt{3}}.y\sqrt{3\left(x-3\right)}}{xy}\le^{Caushy}\dfrac{\dfrac{1}{\sqrt{2}}.x.\dfrac{2+y-2}{2}+\dfrac{1}{\sqrt{3}}.y.\dfrac{3+x-3}{2}}{xy}=\dfrac{\dfrac{1}{2\sqrt{2}}.xy+\dfrac{1}{2\sqrt{3}}.xy}{xy}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}\right)=\dfrac{2\sqrt{3}+3\sqrt{2}}{12}\)
- Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}y-2=2\\x-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\left(n\right)\\x=6\left(n\right)\end{matrix}\right.\)
- Vậy \(MaxA=\dfrac{2\sqrt{3}+3\sqrt{2}}{12}\)
