M= \(\sqrt{x}-1+\sqrt{y-x}\)(đk : \(y\ge x\ge0\)
Áp dụng bđt cosi vs hai số dương có:
\(\sqrt{x}=1.\sqrt{x}\le\frac{x+1}{2}\)
\(\sqrt{y-x}\le\frac{y-x+1}{2}\)
=> \(\sqrt{x}+\sqrt{y-x}\le\frac{x+1}{2}+\frac{y-x+1}{2}\) <=> \(\sqrt{x}-1+\sqrt{y-x}\le\frac{x}{2}+\frac{1}{2}+\frac{y}{2}-\frac{x}{2}+\frac{1}{2}-1\)
<=> \(M\le\frac{y}{2}\) (1)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}1=y-x\\x=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
Từ (1) => M=1 tại y=2,x=1
Vậy maxM=1 <=> x=1,y=2
\(A=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)(đk: \(x,y>0\) ,\(x\ne y\))
=\(\frac{x+y+2\sqrt{xy}-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\) (vì x,y>0)
=\(\frac{x-2\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}+\sqrt{x}-\sqrt{y}\)
= \(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}+\sqrt{x}-\sqrt{y}=\sqrt{x}-\sqrt{y}+\sqrt{x}-\sqrt{y}=2\sqrt{x}-2\sqrt{y}\)
Vậy A= \(2\sqrt{x}-2\sqrt{y}\)
