\(A=3x^3-6x^2+9x-3x^3+2x^2+5x^2-5x=x^2+4x\\ B=\left(x^2+xy+y^2\right)\left(x-y\right)=x^3-y^3\)

\(A=x^2\left(x-y^2\right)-xy\left(1-xy\right)-x^3\\ =x^3-x^2y^2-xy+x^2y^2-x^3\\ =\left(x^3-x^3\right)+\left(-x^2y^2+x^2y^2\right)-xy\\ =-xy\)

\(A=x^2\left(x-y^2\right)-xy\left(1-xy\right)-x^3\)

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

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

\(=-xy\)

Vậy \(A=-xy\)

#\(Toru\)

\(A,xy\left(2x^2-3\right)-x^2\left(5xy+y\right)+x^2y\\ =2x^3y-3xy-5x^3y-x^2y+x^2y\\ =\left(2x^3y-5x^3y\right)+\left(-x^2y+x^2y\right)-3xy\\ =-3x^3y-3xy\)

\(B,3xyz\left(y-2\right)-5yz\left(1-y\right)-8z\left(y^2-3\right)\\ =3xy^2z-6xyz-5yz+5y^2z-8y^2z+24z\\ =3xy^2z-6xyz+\left(5y^2z-8y^2z\right)-5yz+24z\\ =3xy^2z-6xyz-3y^2z-5yz+24z\)

a) (3x2 – 2x2y) : x2 – (2xy2 + x2y) : (1/3 xy)

= (3x3 : x2) + (-2x2y : x2) - [(2x2y : 1/3 xy) +( x2y : 1/3 xy)]

= 3x – 2y – (6y + 3x) = 3x – 2y – 6y – 3x = -8y

\(a,N=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x^4-y^4\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\\ N=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=x^2+y^2\\ b,N=\left(x+y\right)^2-2xy=0-2\cdot1=-2\)

ĐKXĐ: \(x\ne y\)

a) \(N=\dfrac{x^2+y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}:\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=x^2+y^2\)

b) \(x+y=0\Leftrightarrow\left(x+y\right)^2=0\Leftrightarrow x^2+y^2-2xy=0\)

\(\Leftrightarrow N=x^2+y^2=0+2xy=2.1=2\)