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

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

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

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

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

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

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

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

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

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

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

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

Phạm Quốc Cường làm đúng rồi đó

k mình nha

thanks

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

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

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

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

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

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

giúp mk vs bn ơi

P=x^3+3/5x^2y-3xy-3/5x^2y-xy+x^3

=2x^3-4xy

=2*(-2)^3-4*(-2)*1/3

=-16+8/3=-40/3

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

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

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

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

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

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

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

\(=\frac{-2x-1}{x+1}\)(1)

b) Thay \(x=-3,y=2014\)vào (1) ta được:

\(A=\frac{-2.\left(-3\right)-1}{-3+1}=\frac{-5}{2}\)

Vậy \(A=\frac{-5}{2}\)với x=-3 và y=2014