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

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

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

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

\(D=0\)

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

1) Ta có: \(\left\{{}\begin{matrix}2x-y=3\\x+y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=3\\x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-x=-1\end{matrix}\right.\)

Vậy: (x,y)=(1;-1)

2) Ta có: \(A=\dfrac{x+20}{x-4}+\dfrac{2}{\sqrt{x}+2}-\dfrac{6}{\sqrt{x}-2}\)

\(=\dfrac{x+20+2\left(\sqrt{x}-2\right)-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+20+2\sqrt{x}-4-6\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

D

D

a) Ta có: \(\dfrac{x^2}{y^2}:\sqrt{\dfrac{x^2}{y^4}}\)

\(=\dfrac{x^2}{y^2}:\dfrac{x}{y^2}\)

=x

b) Ta có: \(\sqrt{\dfrac{27\left(x-1\right)^2}{12}}+\dfrac{3}{2}-\left(x-2\right)\sqrt{\dfrac{50x^2}{8\left(x-2\right)^2}}\)

\(=\sqrt{\dfrac{9}{4}}\cdot\sqrt{\left(x-1\right)^2}+\dfrac{3}{2}-\left(x-2\right)\cdot\sqrt{\dfrac{25}{4}}\cdot\sqrt{\dfrac{x^2}{\left(x-2\right)^2}}\)

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

\(=\dfrac{3}{2}x-\dfrac{3}{2}+\dfrac{3}{2}-\dfrac{5}{2}\left(x-2\right)\cdot\dfrac{-x}{x-2}\)

\(=\dfrac{3}{2}x+\dfrac{5}{2}\cdot\left(x\right)\)

=4x

em cảm ơn ạ 

a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)

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

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

\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)

\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)

\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)

a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)

do \(x\ge0;y\ge0\)

\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)

1. ĐKXĐ : \(xy>0\)

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

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

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

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

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

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

2. Ta thấy : \(x-\sqrt{xy}+y=x-\dfrac{2.\sqrt{x}.\sqrt{y}}{2}+\dfrac{y}{4}+\dfrac{3y}{4}\)

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

Mà \(\left\{{}\begin{matrix}\left(\sqrt{x}-\dfrac{\sqrt{y}}{2}\right)^2\ge0\\\dfrac{3y}{4}\ge0\end{matrix}\right.\)

\(\Rightarrow x-\sqrt{xy}+y\ge0\)

Lại có : \(\sqrt{xy}\ge0\)

\(\Rightarrow P\ge0\) ( ĐPCM )

a: \(=\dfrac{\left|x+2\right|}{x-1}\)

b: \(=x-2y-\left|x-2y\right|\)\(=\left[{}\begin{matrix}x-2y-x+2y=0\\x-2y+x-2y=2x-4y\end{matrix}\right.\)

c: \(=\dfrac{\left|x+2\right|}{\left(x+2\right)\left(x-2\right)}=\pm\dfrac{1}{x-2}\)

\(a,=\dfrac{x}{y}\cdot\dfrac{\left|x\right|}{y^2}=\dfrac{x^2}{y^3}\\ b,=2y^2\cdot\dfrac{x^2}{\left|2y\right|}=\dfrac{2x^2y^2}{-2y}=-x^2y\)