\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left(z-x\right)\)

\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left[\left(z-y\right)+\left(y-x\right)\right]\)

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

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

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

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

\(=\left(y-x\right)\left(z-y\right)\left[xz\left(z-x\right)+y\left(z-x\right)\left(z+x\right)\right]\)

\(=\left(y-x\right)\left(z-y\right)\left(z-x\right)\left(xy+yz+xz\right)\)

\(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)

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

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

\(=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

chúc bn hc tốt ^^ 

Ta có: \(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)

\(=x\left(y-z\right)\left(y+z\right)+yz^2-x^2y+zx^2-y^2z\)

\(=x\left(y-z\right)\left(y+z\right)-\left(y^2z-yz^2\right)-\left(x^2y-zx^2\right)\)

\(=x\left(y-z\right)\left(y+z\right)-yz\left(y-z\right)-x^2\left(y-z\right)\)

\(=\left(y-z\right)\left(xy+zx-yz-x^2\right)\)

\(=\left(y-z\right)\left[\left(zx-yz\right)-\left(x^2-xy\right)\right]\)

\(=\left(y-z\right)\left[z\left(x-y\right)-x\left(x-y\right)\right]\)

\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz

=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz

=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz

=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3

=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]

=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)

=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]

=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]

=(x+y+z)(x-y-z)(z-x-y)

Câu hỏi của nguyễn khánh linh - Toán lớp 8 - Học toán với OnlineMath

a,Từ giả thiết ta có

(x²+y²+z²)(x+y+z)²+(xy+yz+zx)²

=(x²+y²+z²)(x²+y²+z²+2xy+2yz+2zx)+(xy+yz+zx)²

Đặt x²+y²+z²=a

xy+yz+zx=b

=>(x²+y²+z²)(x²+y²+z²+2xy+2yz+2zx)+(xy+yz+zx)²

=a(a+2b)+b²

=a²+2ab+b²

=(a+b)²

=(x²+y²+z²+xy+yz+zx)²

câu b hơi dài mình gửi sau nhé

Ta có: 2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2-2(x^2+y^2+z^2)(x+y+z)^2+(x+y+z)^4

Gọi x^4+y^4+z^4=a

x^2+y^2+z^2=b

x+y+z=c

=>2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2-2(x^2+y^2+z^2)(x+y+z)^2+(x+y+z)^4=2a-b^2-2bc^2+c^4

=2a-2b^2+b^2-2bc^2+c^4

=2(a-b^2)+(b+c^2)^2

Ta có

2(a-b²)=2[x^4+y^4+z^4-(x^2+y^2+z^2)²]

=2[x^4+y^4+z^4-x^4-y^4-z^4-2x²y²-2y²z²-2z²x²]

=2.(-2)(x²y²+y²z²+z²x²)

=-4(x²y²+y²z²+z²x²)

Lại có

(b+c^2)^2

=[(x^2+y^2+z^2)+(x+y+z)²]²

=[(x^2+y^2+z^2)-(x^2+y^2+z^2)-2(xy+yz+zx)]²

=4(xy+yz+zx)²

=>2(a-b^2)+(b+c^2)^2

=-4(x²y²+y²z²+z²x²)+4(xy+yz+zx)²

=8xyz(x+y+z)

cauu a cua bn Đen đủi .....lm sai r

x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz

=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz

=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz

=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3

=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]

=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)

=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]

=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]

=(x+y+z)(x-y-z)(z-x-y)