\(A=\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\\ =\left(\dfrac{4\sqrt{y}.\left(2-\sqrt{y}\right)+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\dfrac{\sqrt{y}-1-2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\\ =\left(\dfrac{4\sqrt{y}\left(2+\sqrt{y}\right)}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\dfrac{3-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\\ =.\dfrac{4\sqrt{y}.\left(-\sqrt{y}\right)\left(2-\sqrt{y}\right)}{\left(2-\sqrt{y}\right)\left(3-\sqrt{y}\right)}\\ =\dfrac{-4y}{3-\sqrt{y}}\)

Ta có:

\(A=\dfrac{-4y}{3-\sqrt{y}}=-2\Rightarrow-4y=-6+2\sqrt{y}\Rightarrow-4y+4\sqrt{y}-6\sqrt{y}+6=0\\ \Rightarrow-4\sqrt{y}\left(\sqrt{y}-1\right)-6\left(\sqrt{y}-1\right)=0\\ \Rightarrow\left(\sqrt{y}-1\right)\left(-4\sqrt{y}-6\right)=0\Rightarrow\sqrt{y}-1=0\Rightarrow y=1\)

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 $x\neq y; x,y\geq 0$

$\Rightarrow A< 1$

a: \(A=\dfrac{4y-8\sqrt{y}-8y}{y-4}:\dfrac{\sqrt{y}-1-2\sqrt{y}+4}{\sqrt{y}\left(\sqrt{y}-2\right)}\)

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

\(=\dfrac{4y}{\sqrt{y}-3}\)

b: Để A=-2 thì \(4y=-2\sqrt{y}+6\)

=>\(4y+2\sqrt{y}-6=0\)

=>y=1

a:

Sửa đề: \(A=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

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

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

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

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

b: căn xy>0

\(x-\sqrt{xy}+y=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}\sqrt{y}+\dfrac{1}{4}y+\dfrac{3}{4}y\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y>0\)

=>A>0

Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)

5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)