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

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

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

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

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

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

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

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

giúp mk vs bn ơi

a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)

\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)

\(=6x^2-3x+\dfrac{5}{2}\)

b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)

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

\(=2x^2-2y\)

a: =xy(1/3+4-2)=7/3xy

b: =xy^2(-1+3/2+4/3)=(1/3+3/2)xy^2=11/6xy^2

c: =4x^2y^2+2/3x^2y^2-4/3x^2y=-4/3x^2y+14/3x^2y^2

d: =3x^2y^2z+4x^2y^2z-8x^2y^2z=-x^2y^2z

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

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

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

Mình nghĩ là phân tích đa thức 

a)\(3x+2y+xy+6\)

\(=x\left(y+3\right)+2\left(y+3\right)\)

\(=\left(x+2\right)\left(y+3\right)\)

b)\(2x^2+3xy-2y^2-10x-5y+12\)

\(=2x^2+\left(3y-10\right)x-\left(2y^2+5y-12\right)\)

\(=\left[2x+\left(y-4\right)\right]\left(x+2y+3\right)\)

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

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

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

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

\(B=x-y\)\(\left(\text{Đ}K:x+2y\ne0;x+y\ne0\right)\)

Tham khảo nhé~

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

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

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

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

\(=x-y\)