A ) xy(z+y)+yz(y+z)+zx(z+x)

=y.[x(z+y)+z(y+z)]+zx(z+x)

=y.(xz+xy+zy+z²)+zx(z+x)

=y.(xz+z²+xy+zy)+zx(z+x)

=y.[z.(z+x)+y.(z+x)]+zx(z+x)

=y.(z+x)(z+y)+zx(z+x)

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

=(z+x)(yz+y²+zx)

B )xy(x+y)-yz(y+z)-zx(z-x)

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

=y.(x²+xy-zy-z²)-zx(z-x)

=y.(x²-z²+xy-zy)-zx(z-x)

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

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

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

=(x-z)(yx+yz+y²+zx)

=(x-z)(yx+zx+yz+y²)

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

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

b. \(\text{ xy(x+y)-yz(y+z)-xz(z-x) =xy(x+y+z-z)+yz(y+z)+xz(x-z) =xy(x-z)+xy(y+z)+yz(y+z)+xz(x-z) =(x+y)(y+z)(x-z) }\)

a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz

= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]

= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)

= (xy + yz + zx)(x + y + z)

b) Vô câu hỏi tương tự 

a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz

= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]

= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)

= (xy + yz + zx)(x + y + z)

b) tương tự 

xy(x+y)-yz(y+z)-zx(z-x)

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

=y.(x²+xy-zy-z²)-zx.(z-x)

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

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

=y.(z-x)(-x-z-y)-zx.(z-x)

=(z-x)(-xy-zy-y²-zx)

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

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

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

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

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

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

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

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

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

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

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

xy(x+y) + yz(y-z) - zx(z+x)

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

=y(x²+xy+yz-z²)-zx(z+x)

=y(x²-z²+xy+yz)-zx(z+x)

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

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

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

=(x+z)(xy-yz+y²-xyz)

a/ \(\left(x-y\right)\left(z-x\right)\left(z-y\right)\)

b/ \(\left(1-y\right)\left(y-x\right)\)

a. \(\left(x-y\right)\left(z-x\right)\left(z-y\right)\)

b. \(\left(1-y\right)\left(y-x\right)\)

a. 

b. 

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

\(=xy^2-xz^2+yz^2-x^2y+zx^2-zy^2\)

\(=xy^2-xz^2+yz^2-x^2y+zx^2-zy^2-xyz+xyz\)

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

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

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

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

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

phân tích đa thức thành nhân tử đặt biến phụ

(x2 + y2 + z2)(x + y + z)2 + (xy + yz + zx)2

(x2 + y2 + z2)(x + y + z)2 + (xy + yz +zx)2

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

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

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

= (x2 + y2 + z2 + xy + yz + zx)2

Đảm bảo ko phân tích tiếp đc nữa đâu ^^, đây tuy ko phải cách đặt biến phụ nhưng cách này chắc ngắn hơn cách đặt biến phụ.

  bởi Bùi Xuân Chiến