Question

已知xyz=1,x+y+z=2,x²+y²+z²=3,求1/(xy+z-1)+1/(yz+

Answer 1

(x+y+z)^2-(x^2+y^2+z*2)=2xy+2yz+2zx=2^2-3=1

xy+yz+zx=1/2

(x-1)(y-1)(z-1)=xyz-xy-yz-zx+x+y+z-1=1-1/2+2-1=3/2

1/(xy+z-1)+1/(yz+x-1)+1/(zx+y-1)

=1/(xy+1-x-y)+1/(yz+1-y-z)+1/(zx+1-z-x)

=1/[(x-1)(y-1)]+1/[(y-1)(z-1)]+1/[(z-1)(x-1)]

=(z-1+x-1+y-1)/[(x-1)(y-1)(z-1)]

=(2-3)/(3/2)=-2/3