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16 tháng 10 2017

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

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

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

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

\(A=\left(y-z\right)\left[\left(x^2y^2-x^3y\right)+\left(x^2yz-x^3z\right)+\left(x^2z^2-y^2z^2\right)\right]\)

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

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

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

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

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

28 tháng 10 2015

y(x+y)+yz(y+z)+xz(x+z)+2xyz 

= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 

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

= xy(x + y) + z(x + y + z)(y + x) 

= (x + y)(xy + zx + zy + z²) 

= (x + y)[x(y + z) + z(y + z)] 

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

28 tháng 10 2015

Monkey D.Luffy copy ở đâu mà hay z

9 tháng 8 2015

b)x(y+z)2+y(z+x)2+z(x+y)2-4xyz

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

=x(y2+2yz+z2-2yz)+y(x2+z2+2xz-2xz)+z(x+y)2

=x(y2+z2)+y(x2+z2)+z(x+y)2

=xy2+xz2+x2y+yz2+(xz+yz)(x+y)

=xy(x+y)+z2(x+y)+(xz+yz)(x+y)

=(x+y)(xy+z2+xz+yz)

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

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

9 tháng 8 2015

a)x(y2-z2)+y(z2-x2)+z(x2-y2)

=x(y-z)(y+z)+yz2-x2y+x2z-y2z

=(y-z)(xy+xz)-x2(y-z)-yz(y-z)

=(y-z)(xy+xz-x2-yz)

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

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

nhấn vào đây nhé có 2 cách làm: Chuyên đề Bồi dưỡng học sinh giỏi - Phân tích đa thức thành nhân tử - Giáo Án, Bài Giảng

t i c k mk!! 536546456545576768978045362546115346456575676868784675462552

27 tháng 10 2019

Câu hỏi của Kim Lê Khánh Vy - Toán lớp 8 - Học toán với OnlineMath

14 tháng 8 2020

Ta có :

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

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

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

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

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

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

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

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

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

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

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

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

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

14 tháng 8 2020

Ta có :

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

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

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

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

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

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

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

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

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

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

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

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

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

13 tháng 8 2020

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

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

\(=xy^2\left(1-1\right)+yz^2\left(1-1\right)+zx^2\left(1-1\right)\)

\(=\left(xy^2+yz^2+zx^2\right).0\left(=0\right)\)

13 tháng 6 2015

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

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

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

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

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

\(=y^2\left(z-x\right)\left(-yz-xy-z^2-zx-x^2\right)-z^2x^2\left(z-x\right)=\left(z-x\right)\left(-y^3z-xy^2-z^2y^2-xyz-x^2y^2-z^2x^2\right)\)

đến đây coi như là thành nhân tử rồi nha. em muốn gọn thì ráng ngồi nghĩ rồi tách nha. chỉ cần nhóm mấy cái có ngoặc giống nhau là đc. k khó đâu. chịu khó nghĩ để rèn luyện nha

c) \(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)

\(\left(9a^3-6a^2\right)+\left(6a^2-4a\right)+\left(-9a+6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)

d) em sửa đề đi. đề sai rồi. đồng nhất hệ số phải có dấu bằng nha.

có gì liên hệ chị. đúng nha ;)

21 tháng 8 2021

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

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

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

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

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

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

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

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