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

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

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

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

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

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

\(=\dfrac{1}{x-y}\)

Rút gọn :
b ) \(\frac{x^2+3xy+2y^2}{x^2+2x^2y-xy^2-2y^2}\)

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

\(=\frac{x\left(x+4\right)+2y\left(x+y\right)}{x\left(x^2-y^2\right)+2y\left(x^2-y^2\right)}\)

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

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

\(=\frac{1}{x-y}\)

B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

giúp mk vs bn ơi

\(\frac{x^2-x-6}{x^2+7x+10}\)

\(=\frac{x^2-3x+2x-6}{x^2+5x+2x+10}=\frac{x.\left(x-3\right)+2.\left(x-3\right)}{x.\left(x+5\right)+2.\left(x+5\right)}\)

\(=\frac{\left(x+2\right).\left(x-3\right)}{\left(x+2\right).\left(x+5\right)}=\frac{x-3}{x+5}\)