早教吧作业答案频道 -->数学-->
1.xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)2.(y+z)(z+x)(x+y)+xyz是因式分解
题目详情
1.xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
2.(y+z)(z+x)(x+y)+xyz
是因式分解
2.(y+z)(z+x)(x+y)+xyz
是因式分解
▼优质解答
答案和解析
1.xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
=xy(x^2-z^2+z^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
=xy(x^2-z^2)+xy(z^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
=(z^2-x^2)(zx-xy)+(y^2-z^2)(yz-xy)
=(z-x)(z-y)(x(z+x)-y(y+z))
=(z-x)(z-y)(z(x-y)+(x+y)(x-y))
=(z-x)(z-y)(x-y)(x+y+z)
2.(y+z)(z+x)(x+y)+xyz
=(x+y+z-x)(z+x)(x+y)+xyz
=(x+y+z)(z+x)(x+y)-x(z+x)(x+y)+xyz
=(x+y+z)(z+x)(x+y)-x^2z-x^3-x^2y
=(x+y+z)(z+x)(x+y)-x^2(x+y+z)
=(x+y+z)(xz+zy+xy)
=xy(x^2-z^2+z^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
=xy(x^2-z^2)+xy(z^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
=(z^2-x^2)(zx-xy)+(y^2-z^2)(yz-xy)
=(z-x)(z-y)(x(z+x)-y(y+z))
=(z-x)(z-y)(z(x-y)+(x+y)(x-y))
=(z-x)(z-y)(x-y)(x+y+z)
2.(y+z)(z+x)(x+y)+xyz
=(x+y+z-x)(z+x)(x+y)+xyz
=(x+y+z)(z+x)(x+y)-x(z+x)(x+y)+xyz
=(x+y+z)(z+x)(x+y)-x^2z-x^3-x^2y
=(x+y+z)(z+x)(x+y)-x^2(x+y+z)
=(x+y+z)(xz+zy+xy)
看了1.xy(x^2-y^2)+y...的网友还看了以下:
1.已知x+y+z=6,xy+yz+xz=7,则x²+y²+z²=?2.(2a+2b+1)(2a+ 2020-04-07 …
化简(2x-y-z/x^2-xy-xz+yz)+(2y-x-z/y^2-xy-yz+xz)+(2x 2020-05-16 …
设x、y、z为实数,且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+ 2020-06-12 …
1)2x^3-13x^2+25x-142)(x+1)^4+(3+x)^4-2723)xy(xy+1 2020-07-18 …
已知xyz=1,x+y+z=2,x^2+y^2+z^2=16.求代数式1\(xy+2z)+1\(y 2020-07-31 …
附加题:(y-z)2+(x-y)2+(z-x)2=(y+z-2x)2+(z+x-2y)2+(x+y- 2020-10-30 …
已知三个数x,y,z满足xy/(x+y)=-2,yz/(y+z)=3/4,zx/(z+已知三个数x, 2020-11-01 …
初二数学题(1)(-a^2b/c^3)^2×(bc/-a^2)^3÷(b/ac)^4(2)(x+y/ 2020-11-01 …
1.已知1=xy/(x+y),2=yz/(y+z),3=zx/(z+x),则x+y+z=?2..当x 2020-11-01 …
对称式的问题1.证明:3个变数多项式f(x,y,z)=x(y-z)^2+y(z-x)^2+z(x-y 2020-11-07 …