Xem ở đây nhé Phân tích đa thức thành nhân tử

4x(x+y)(x+y+z)(x+z)+y²z²=4(x²+xy+xz)(x²+xy+xz+yz)+y²z²=4(x²+xy+xz)²+4yz(x²+xy+xz)+y²z²=(2(x²+xy+xz)+yz)²=(2x²+2xy+2xz+yz)

\(2xyz+x^2y+xy^2+x^2z+xz^2+y^2z+yz^2\)

\(=x^2\left(y+z\right)+yz\left(y+z\right)+x\left(y^2+z^3\right)+2xyz\)

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

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

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

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

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

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

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

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

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

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

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

Mình nghĩ bạn ghi đề sai, đề đúng theo mình là:

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

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

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

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

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

\(=\left(x-y\right)\left(x-z\text{ }\right)\text{[}y^2.\left(x+z\right)-z^2\left(x+y\right)\text{]}\)

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

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

\(=\left(x-y\right)\left(z-x\right)\text{[}x.\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\text{]}\)

\(=\left(x-y\right)\left(x-z\right)\left(y-z\right)\left(xy+x\text{z}+yz\right)\)

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

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

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

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

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

tuổi con HN là :

50 : ( 1 + 4 ) = 10 ( tuổi )

tuổi bố HN là :

50 - 10 = 40 ( tuổi )

hiệu của hai bố con ko thay đổi nên hiệu vẫn là 30 tuổi

ta có sơ đồ : bố : |----|----|----|

                  con : |----| hiệu 30 tuổi

tuổi con khi đó là :

 30 : ( 3 - 1 ) = 15 ( tuổi )

số năm mà bố gấp 3 tuổi con là :

 15 - 10 = 5 ( năm )

       ĐS : 5 năm

mình nha

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

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

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

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

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

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

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

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

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

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

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

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

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