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

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

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

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

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

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

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

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

giúp mk vs bn ơi

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

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

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

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

1. G= 3x²y - 2xy² + x³y³ + 3xy² - 2x²y - 2x³y³

G = x²y + xy² - x³y³ = xy (x + y -x²y²)  . Khi x= -2 . y=4 ta có G= -2*4( -2 + 4 - (-2)² * 4² ) = 496

a. B+A =( -2x² + xy +2y² -5x +2y - 3) + ( x² -3xy -y² +2x -3y +1)= -x² - 2xy + y² -3x -y -2 

A-B= -( -2x² +xy + 2y² -5x +2y -3) + ( x² -3xy -y² + 2x -3y +1) = 3x² -4xy -3y² +7x -5y +4

Tại x = -1, y =2

A= (-1)² -3*(-1)*2 -2² +2*(-1) -3*2 +1 = -4

B= -2*(-1)² + (-1)*2 + 2*2² -5*(-1) + 2*2 -3 = 10

Câu 2 bạn xem lại đề nhé. hình như bạn nhầm đề!

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

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

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

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