已知xyz=1求1/（xy+x+1)+1/(yz+y+1)+1/(xz+z+1)

Question

已知xyz=1求1/（xy+x+1)+1/(yz+y+1)+1/(xz+z+1)

Answer 1

主要是利用好xyz=1这个条件
易有x、y、z均不为零
原式=xyz/(xy+x+xyz)+1/(yz+y+1)+1/(xz+z+1)////第一项可化简,同除以x
=yz/(yz+y+1)+1/(yz+y+1)+1/(xz+z+1)
=（yz+1）/(yz+y+1)+1/(xz+z+1)
=(yz+xyz)/(yz+y+xyz)+1/(xz+z+1)
=(yz+xZ)/(xz+z+1)+1/(xz+z+1)
=(yz+xz+1)/(xz+z+1)=(x+y+xy)/(xy+x+1)