\(\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)}+\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{x}\right)}+\)\(\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{z}-\sqrt{y}\right)}\)

\(=-\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{z}-\sqrt{x}\right)}-\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(-\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\frac{-x\left(\sqrt{y}-\sqrt{z}\right)-y\left(\sqrt{z}-\sqrt{x}\right)-z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{-x\sqrt{y}+x\sqrt{z}-y\sqrt{z}+y\sqrt{x}-z\sqrt{x}+z\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{-\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{z}\left(x-y\right)-z\left(\sqrt{x}-y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{-\sqrt{xy}+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{-\sqrt{xy}+\sqrt{xz}+\sqrt{yz}-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{\sqrt{y}\left(\sqrt{z}-\sqrt{x}\right)-\sqrt{z}\left(\sqrt{z}-\sqrt{x}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(\left|\frac{x+y}{2}-\sqrt{xy}\right|+\left|\frac{x+y}{2}+\sqrt{xy}\right|=\left|\frac{x+2\sqrt{xy}+y}{2}\right|+\left|\frac{x-2\sqrt{xy}+y}{2}\right|\)

=\(\left|\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2}\right|+\left|\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{2}\right|\) (*)

Có \(\left(\sqrt{x}+\sqrt{y}\right)^2\ge0\Rightarrow\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2}\ge0\)

\(\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\Rightarrow\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{2}\ge0\)

\(\Rightarrow\) (*) \(\Leftrightarrow\) \(\frac{x+2\sqrt{xy}+y+x-2\sqrt{xy}+y}{2}=\frac{2\left(x+y\right)}{2}=x+y=\left|x\right|+\left|y\right|\) ( vì x ; y >0)

Với x,y < 0 , đẳng thức trên sai ngay từ bước biến đổi (*) , vì x,y <0 thì \(\sqrt{x}\) và \(\sqrt{y}\) không xác định

Ta có:

\(VT=\left(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\right)\\ =\left(\frac{\sqrt{x}\cdot\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\right)\\ =\left(\frac{\sqrt{xy}\left[\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2\right]}{\sqrt{xy}}\right)\\ =x-y=VP\left(đpcm\right)\)

Vậy với x>0, y>0 ta có đpcm

\(\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)= x-y

=\(\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)

= \(x-y=x-y\)

Sai đề

tớ ra kết quả là 2+\(\frac{5\sqrt{xy}}{x-\sqrt{xy}+y}\) mà thấy số xấu quá :(

ĐKXĐ:

\(P=\left(\frac{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{x\sqrt{x}-y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{\left(\sqrt{x}+\sqrt{y}\right)}{x-\sqrt{xy}+y}\)

\(=\left(\frac{x\sqrt{y}-y\sqrt{x}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\left(\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\right)\)

\(=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)}.\frac{1}{\left(x-\sqrt{xy}+y\right)}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

\(P=\frac{\sqrt{xy}}{x-\sqrt{xy}+\frac{y}{4}+\frac{3y}{4}}=\frac{\sqrt{xy}}{\left(\sqrt{x}-\frac{\sqrt{y}}{2}\right)^2+\frac{3y}{4}}\)

Do \(\left\{{}\begin{matrix}\sqrt{xy}\ge0\\y\ge0\end{matrix}\right.\) \(\Rightarrow P\ge0\) \(\forall x;y\)