Ta phân tích mẫu:

\(x^3+2x^2y-xy^2-2y^3\)

\(=x^3+3x^2y+2xy^2-x^2y-3xy^2-2y^3\)

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

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

Thay vào ta có:

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

Vậy ta có điều phải chứng minh

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

\(DK\hept{\begin{cases}x^3+2x^2y-xy^2-2y^3\ne0\\x-y\ne0\end{cases}}\)

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

\(\Leftrightarrow x^3+3x^2y+2xy^2-x^2y-3xy^2-2y^3=x^3+2x^2y-xy^2-2y^3\)

\(\Leftrightarrow x^2y=0\)\(\Rightarrow ko.dung.\)

?????????

Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)

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

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

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

Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

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

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

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

\(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

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

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