Question

cho x,y thỏa mãn \(3\left(x\sqrt{y-9}+y\sqrt{x-9}\right)=xy\)

Tính \(p=\left(x-17\right)^{2018}+\left(y-19\right)^{^{ }2019}\)

Answer 1

\(3\left(x\sqrt{y-9}+y\sqrt{x-9}\right)=xy\Leftrightarrow\dfrac{3x\sqrt{y-9}+3y\sqrt{x-9}}{xy}=1\)

\(\Leftrightarrow\dfrac{3\sqrt{x-9}}{x}+\dfrac{3\sqrt{y-9}}{y}=1\)

Áp dụng BĐT \(a.b\le\dfrac{a^2+b^2}{2}\) ta có:

\(\dfrac{3\sqrt{x-9}}{x}+\dfrac{3\sqrt{y-9}}{y}\le\dfrac{3^2+x-9}{2x}+\dfrac{3^2+y-9}{2y}=\dfrac{1}{2}+\dfrac{1}{2}=1\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-9}=3\\\sqrt{y-9}=3\end{matrix}\right.\) \(\Rightarrow x=y=18\)

Thay vào P ta được:

\(P=\left(18-17\right)^{2018}+\left(18-19\right)^{2019}=1^{2018}+\left(-1\right)^{2019}=1-1=0\)