\(=\frac{x^2+xy+y^2}{x+y}.\left(\frac{1}{\left(x-y\right)x}-\frac{3y^2}{x\left(x^3-y^3\right)}-\frac{y}{x\left(x^2+xy+y^2\right)}\right)\)

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

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

\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)

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

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

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

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)

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

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\), ta có:

\(A=\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\times\dfrac{2}{a+b}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\)\(\times\dfrac{a^3+ab^2+a^2b+b^3}{ab^3+a^3b}\)

\(=\left(\dfrac{b+a}{ab}\times\dfrac{2}{a+b}+\dfrac{b^2+a^2}{a^2b^2}\right)\)\(\times\dfrac{a^2\left(a+b\right)+b^2\left(a+b\right)}{ab\left(a^2+b^2\right)}\)

\(=\dfrac{2ab+b^2+a^2}{a^2b^2}\times\dfrac{\left(a+b\right)\left(a^2+b^2\right)}{ab\left(b^2+a^2\right)}\)

\(=\dfrac{\left(a+b\right)^3}{a^3b^3}\)

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

\(\left(\dfrac{x^2}{x+y}+y\right).\left(\dfrac{1}{x^2-xy}-\dfrac{3y^3}{x^4-xy^3}-\dfrac{y}{x^3+x^2y+xy^2}\right)\)

\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{1}{x\left(x-y\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{y}{x\left(x^2+xy+y^2\right)}\right)\)\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{x^2+xy+y^2}{x\left(x^3-y^3\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{xy-y^2}{x\left(x^3-y^3\right)}\right)\)

\(=\dfrac{x\left(x^3-y^3\right)}{x^3-xy^2}.\dfrac{x^2+xy+y^2-3y^2-xy+y^2}{x\left(x^3-y^3\right)}\\ =\dfrac{x^2-y^2}{x\left(x^2-y^2\right)}=\dfrac{1}{x}\)

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