Question

tìm Max của \(\frac{x\sqrt{y-2}+y\sqrt{x-3}}{xy}\) với x>=3;y>=2

Answer 1

\(\frac{x\sqrt{y-2}+y\sqrt{x-3}}{xy}=\frac{\sqrt{y-2}}{y}+\frac{\sqrt{x-3}}{x}\)

Áp dụng BĐT Cauchy ta có : \(\frac{\sqrt{\left(y-2\right).2}}{\sqrt{2}y}\le\frac{y-2+2}{2\sqrt{2}y}=\frac{1}{2\sqrt{2}}\)

\(\frac{\sqrt{\left(x-3\right).3}}{\sqrt{3}x}\le\frac{x-3+3}{2\sqrt{3}x}=\frac{1}{2\sqrt{3}}\)

Vậy \(\frac{x\sqrt{y-2}+y\sqrt{x-3}}{xy}\le\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x=6\\y=4\end{cases}}\)

Vậy ..................................

Answer 2

tks :)