由已知条件：x+y+z=2x^2+y^2+z^2=3所以xy+yz+zx=(1/2)[(x+y+z)^2-(x^2+y^2+z^2)]=1/2又因为左式第一项1/(xy+z-1)=1/[xy+(2-x-y)-1]=1/[(x-1)(y-1)]同理1/(yz+x-1)=1/[(y-1)(z-1)]1/(zx+y-1)=1/[(z-1)(x-1)]三式相加...