Câu hỏi của trang souciu

Question

phân tích đa thức thành nhân tử $\left(x+y+z\right)^3-x^3-y^3-z^3$$x^4+2010x^2+2009x+2010$

Thuận Quốc · Accepted Answer

x4+2010x2+2009x+2010=x4-x+2010x2+2010x+2010=x.(x3-1)+2010.(x2+x+1)=x.(x-1)(x2+x+1)+2010.(x2+x+1)=(x2+x+1)(x2-x+2010)Câu trả lời: x4+2010x2+2009x+2010=x4-x+2010x2+2010x+2010=x.(x3-1)+2010.(x2+x+1)=x.(x-1)(x2+x+1)+2010.(x2+x+1)=(x2+x+1)(x2-x+2010)

Thuận Quốc · Answer

(x+y+z)3-x3-y3-z3=(x+y+z-x)[(x+y+z)2+(x+y+z).x+x2]-(y+z)(y2-yz+z2)=(y+z)(x2+y2+z2+2xy+2yz+2zx+x2+xy+zx+x2)-(y+z)(y2-yz+z2)=(y+z)(3x2+y2+z2+3xy+2yz+3zx)-(y+z)(y2-yz+z2)=(y+z)(3x2+y2+z2+3xy+2yz+3zx-y2+yz-z2)=(y+z)(3x2+3yz+3xy+3zx)=3.(y+z)(x2+xy+yz+zx)=3.(y+z)[x.(x+y)+z.(x+y)=3.(y+z)(x+y)(x+z)Câu trả lời: (x+y+z)3-x3-y3-z3=(x+y+z-x)[(x+y+z)2+(x+y+z).x+x2]-(y+z)(y2-yz+z2)=(y+z)(x2+y2+z2+2xy+2yz+2zx+x2+xy+zx+x2)-(y+z)(y2-yz+z2)=(y+z)(3x2+y2+z2+3xy+2yz+3zx)-(y+z)(y2-yz+z2)=(y+z)(3x2+y2+z2+3xy+2yz+3zx-y2+yz-z2)=(y+z)(3x2+3yz+3xy+3zx)=3.(y+z)(x2+xy+yz+zx)=3.(y+z)[x.(x+y)+z.(x+y)=3.(y+z)(x+y)(x+z)

Hao Khi Viet Nam · Answer

=x3+y3+z3+3(x+y)(x+z)(y+z)-x3-y3-z3=3(x+y)(x+z)(y+z)-x3-y3-z3Câu trả lời: =x3+y3+z3+3(x+y)(x+z)(y+z)-x3-y3-z3=3(x+y)(x+z)(y+z)-x3-y3-z3

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)