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

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) 

Lời giải:

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

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

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

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

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

\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{y-x}+\dfrac{1}{x+2\sqrt{x}\sqrt{y}+y}\right)-2x\) (với \(x\ne y,x,y\ge0\))

\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}+\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)

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

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

\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{2\sqrt{y}}{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)-2x\)

\(P=\dfrac{4\sqrt{xy}}{x-y}\cdot\dfrac{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{2\sqrt{y}}-2x\)

\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\cdot2\sqrt{y}}-2x\)

\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\cdot2\sqrt{y}}-2x\)

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

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

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

\(P=\dfrac{2y\sqrt{x}-4x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)

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

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

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

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

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

c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)

\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)

\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)

\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)

d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)

\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)

\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)

\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)

\(D=0\)

Lời giải:

a) ĐK: $x\geq 0; y\geq 0; x\neq y$

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

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

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

b) \(1-A=\frac{(\sqrt{x}-\sqrt{y})^2}{x-\sqrt{xy}+y}>0\) với mọi $x\neq y; x,y\geq 0$

$\Rightarrow A< 1$