xy(x-y)+yz(y-z)+zx(z-x)=y[x(x-y)+z(y-z)]+zx(z-x)=y[x^2-xy+zy-z^2]+zx(z-x)=y[x^2-z^2+zy-xy]+zx(z-x)=y[(x-z)(x+z)+y(z-x)]+zx(z-x)=(z-x)[-y(x+z)+y^2+zx]=(z-x)(y^2-xy-yz+zx)=(z-x)[y(y-z)-x(y-z)]=(z-x)(y-z...
xy(x-y)+yz(y-z)+zx(z-x)因式分解
1个回答
相关问题
-
化简x^2-yz/[x^2-(y+z)x+yz]+y^2-zx/[y^2-(z+x)y+zx]+z^2-xy/[z^2-
-
x/3=y/2=z/5,(xy+yz+zx)/(x²+y²+z²)=
-
因式分解:x2y+y2z+z2x+xy2+yz2+zx2+3xyz
-
设x,y,z≥0,x+y+z=3,证明:√x+√y+√z≥xy+yz+zx
-
设X Y Z属于R,求证x²+y²+z²≥xy+yz+zx
-
方程组xy/x+y=2 yz/y+z=4 zx/x+z=5 求x,y,z
-
分式题:xy=x+y,yz=2(y+z),zx=3(z+x),求xyz/(xy+yz+xz)
-
已知三个数x,y,z满足xy/x+y=-2,yz/y+z=3/4,zx/z+x=-3/4,则xyz/(xy+yz+zx)
-
xy+yz+zx=1,x,y,z>=0
-
已知x+y+z=1,xy+yz+zx=xyz,求证:(1-x)(x+yz)=0 ,(1-y)(y+zx)=0,(1-z)