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

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

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

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

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

2: \(\left(x^2-y^2\right)\cdot C=-8\)

=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)

=>\(\left(x-y\right)^3=-8\)

=>x-y=-2

=>x=y-2

\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)

\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)

\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)

\(=\left(y-1\right)\left(-4y+4\right)+4xy\)

\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)

\(=-4y^2+8y-4+4y^2-8y\)
=-4

\(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)

\(=\left(\dfrac{x-1-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)

\(=\left(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

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

\(=\dfrac{\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

Ta có: \(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)

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

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

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

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

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

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

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

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

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

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

\(=-1\)

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

\(A\ge\dfrac{2xy}{2xy}+2\sqrt{\left(\dfrac{x^2+y^2}{2xy}\right)\left(\dfrac{2xy}{x^2+y^2}\right)}=3\)

Dấu "=" xảy ra khi \(x=y\)

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

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

\(B\ge2\sqrt{\dfrac{\left(x+y\right)^2.4xy}{4xy.\left(x+y\right)^2}}+\dfrac{3}{4}.\dfrac{4xy}{xy}-4=1\)

\(B_{min}=1\) khi \(x=y\)

\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)

\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)

Có \(\frac{2}{x^2-y^2}\sqrt{\frac{9\left(x+2xy+y\right)}{4}}\) 

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

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

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

=\(\frac{2.3.\left(x+y\right)}{\left(x+y\right)\left(x-y\right).2}\) =\(\frac{3}{x-y}\)

\(B=\dfrac{2}{xy}\div\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2-\dfrac{x^2+y^2}{\left(x-y\right)^2}\)

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

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

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

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

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

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

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

\(B=\dfrac{-\left(x-y\right)^2}{\left(x-y\right)^2}=-1\)

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

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

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

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

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

\(=\dfrac{\left(x-y\right)^2}{x+y}:\dfrac{\left(x-y\right)^2}{x^2-y^2}=\dfrac{\left(x-y\right)^2}{x+y}.\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^2}=x-y\)