a) \(B=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x,y\ge0;x\ne y\right)\)

\(B=\left[\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{x-y}\right]:\dfrac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(B=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)

\(B=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)

\(B=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)

\(B=\dfrac{\sqrt{xy}}{x+\sqrt{xy}+y}\)

b) Xét tử: 

\(\sqrt{xy}\ge0\forall x,y\) (xác định) (1) 

Xét mẫu: 

\(x+\sqrt{xy}+y\)

\(=\left(\sqrt{x}\right)^2+2\cdot\dfrac{1}{2}\sqrt{y}\cdot\sqrt{x}+\left(\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)

\(=\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)

Mà: \(\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2\ge0\forall x,y\) (xác định), còn: \(\dfrac{3}{4}y\ge0\) vì theo đkxđ thì \(y\ge0\) (2) 

Từ (1) và (2) ⇒ B luôn không âm với mọi x,y (\(B\ge0\)) (đpcm) 

Ta có: \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt{xy}}\)

\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}\right)+\sqrt{y}\left(x-\sqrt{xy}\right)}{x^2-xy}\)

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

\(=\dfrac{2\sqrt{y}}{x-y}=\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(\Rightarrow A=\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}.\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\dfrac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\dfrac{\sqrt{x}+\sqrt{y}-1+1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)

Bạn cần làm gì với biểu thức này?
 

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

\(=\dfrac{x\sqrt{x}\left(1-y\right)+x\left(1-y\sqrt{y}\right)-y\left(1-\sqrt{y}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\dfrac{\left(1-\sqrt{y}\right)\left[x\sqrt{x}\left(1+\sqrt{y}\right)+x+x\sqrt{y}+xy-y\right]}{\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

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

\(=\dfrac{x\left(\sqrt{x}+1\right)+x\sqrt{y}\left(\sqrt{x}+1\right)+y\left(x-1\right)}{\left(1+\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\dfrac{x+x\sqrt{y}+y\sqrt{x}-y}{\sqrt{x}+\sqrt{y}}=\sqrt{x}-\sqrt{y}+\sqrt{xy}\)

Để A=2 thì x=2; y=2

a: \(N=\dfrac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)-x\sqrt{x}+y\sqrt{y}}{x-y}\right)\)

\(=\dfrac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}:\dfrac{x\sqrt{x}+x\sqrt{y}-y\sqrt{x}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}{x-y}\)

\(=\dfrac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{x-y}{x\sqrt{y}-y\sqrt{x}}\)

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

\(=\dfrac{x-\sqrt{xy}+y}{\sqrt{xy}}\)

b: \(N-1=\dfrac{x-2\sqrt{xy}+y}{\sqrt{xy}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}}>0\)

=>N>1