Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-x-y+x\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-x\right)-y^2z^2\left(y-x\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(y-x\right)\left(x-z\right)\left(x+z\right)+z^2\left(z-x\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y^2x+y^2z-z^2y-z^2x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y-z\right)\left(xy+yz+zx\right)\)
b) Dùng phương pháp đặt ẩn phụ:
Đặt y - x = a; z - y = b suy ra \(a+b=y-x+z-y=z-x\)
\(x^2y^2a+y^2z^2b-z^2x^2\left(a+b\right)=\left(x^2y^2a-z^2x^2a\right)+\left(y^2z^2b-z^2x^2b\right)\)
\(=x^2a\left(y^2-z^2\right)+z^2b\left(y^2-x^2\right)=x^2\left(y-x\right)\left(y-z\right)\left(y+z\right)+z^2\left(z-y\right)\left(y-x\right)\left(x+y\right)\)
\(=x^2\left(y-x\right)\left(y-z\right)\left(y+z\right)-z^2\left(y-z\right)\left(y-x\right)\left(x+y\right)\)
\(=\left(y-x\right)\left(y-z\right)\left[x^2\left(y+z\right)-z^2\left(x+y\right)\right]\)
\(=\left(y-x\right)\left(y-z\right)\left(x^2y+x^2z-z^2x-z^2y\right)\)
\(=\left(y-x\right)\left(y-z\right)\left[y\left(x^2-z^2\right)+xz\left(x-z\right)\right]\)
\(=\left(y-x\right)\left(y-z\right)\left[y\left(x-z\right)\left(x+z\right)+xz\left(x-z\right)\right]\)
\(=\left(y-x\right)\left(y-z\right)\left(x-z\right)\left(xy+yz+zx\right)\)
\(a)\)\(\left(x^2+y^2-5\right)^2-4x^2y^2-16xy-16\)
\(=\)\(\left(x^2+y^2-5\right)^2-\left(4x^2y^2+16xy+16\right)\)
\(=\)\(\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)
\(=\)\(\left(x^2-2xy+y^2-5+4\right)\left(x^2+2xy+y^2-5-4\right)\)
\(=\)\(\left[\left(x-y\right)^2-1\right].\left[\left(x+y\right)^2-9\right]\)
\(=\)\(\left(x-y-1\right)\left(x-y+1\right)\left(x+y-9\right)\left(x+y+9\right)\)
Chúc bạn học tốt ~
\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)=x^2\left(y-z\right)-y^2\left[\left(y-z\right)+\left(x-y\right)\right]+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(y-z\right)-y^2\left(x-y\right)+z^2\left(x-y\right)\)
\(=\left(x^2-y^2\right)\left(y-z\right)-\left(y^2-z^2\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(y-z\right)-\left(y-z\right)\left(y+z\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x+y-y-z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
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
Câu hỏi của Kim Lê Khánh Vy - Toán lớp 8 - Học toán với OnlineMath
Phân tích các đa thức sau thành nhân tử:
a) x(y2-z2)+y(z2-x2)+z(x2-y2)
b) x(y+z)2+y(z+x)2+z(x+y)2-4xyz
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)
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)
Ta có :
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)^2.\left(x-y\right)+\left(y+z\right).\left(y^2-x^2+x^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2+z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2\right)-\left(y+z\right)\left(z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x^2-y^2\right)\left(x+y-y-z\right)-\left(z^2-x^2\right).\left(y+z-z-x\right)\)
\(=\left(x^2-y^2\right).\left(x-z\right)-\left(z^2-x^2\right).\left(y-x\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x-z\right)+\left(z-x\right)\left(z+x\right)\left(x-y\right)\)
\(=\left(x-y\right).\left[\left(x+y\right)\left(x-z\right)+\left(z-x\right).\left(x+z\right)\right]\)
\(=\left(x-y\right)\left(x^2-zx+xy-yz+zx+z^2-x^2-xz\right)\)
\(=\left(x-y\right)\left(z^2-zx+xy-yz\right)\)
\(=\left(x-y\right)\left[z.\left(z-x\right)-y.\left(z-x\right)\right]\)
\(=\left(x-y\right)\left(z-y\right)\left(z-x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
Ta có :
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)^2.\left(x-y\right)+\left(y+z\right).\left(y^2-x^2+x^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2+z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2\right)-\left(y+z\right)\left(z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x^2-y^2\right)\left(x+y-y-z\right)-\left(z^2-x^2\right).\left(y+z-z-x\right)\)
\(=\left(x^2-y^2\right).\left(x-z\right)-\left(z^2-x^2\right).\left(y-x\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x-z\right)+\left(z-x\right)\left(z+x\right)\left(x-y\right)\)
\(=\left(x-y\right).\left[\left(x+y\right)\left(x-z\right)+\left(z-x\right).\left(x+z\right)\right]\)
\(=\left(x-y\right)\left(x^2-zx+xy-yz+zx+z^2-x^2-xz\right)\)
\(=\left(x-y\right)\left(z^2-zx+xy-yz\right)\)
\(=\left(x-y\right)\left[z.\left(z-x\right)-y.\left(z-x\right)\right]\)
\(=\left(x-y\right)\left(z-y\right)\left(z-x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=-xy^2+yx^2-yz^2+zy^2-xz^2+zx^2\)
\(=xy^2\left(1-1\right)+yz^2\left(1-1\right)+zx^2\left(1-1\right)\)
\(=\left(xy^2+yz^2+zx^2\right).0\left(=0\right)\)