设xyz=1,求(1/1+x+xy)+(1/1+y+yz)+(1/1+z+zx)的值设xyz=1,求(x/1+x+xy)+(y/1+y+yz)+(z/1+z+zx)的值抱歉弄错了

Question

设xyz=1,求(1/1+x+xy)+(1/1+y+yz)+(1/1+z+zx)的值
设xyz=1,求(x/1+x+xy)+(y/1+y+yz)+(z/1+z+zx)的值抱歉弄错了

Answer 1

1＋x＋xy分之1＋（1＋y＋yz）分之1＋（1＋z＋xz）分之1
=1/(1+x+xy)+1/(1+y+yz)+1/(1+z+xz)
=1/(xyz+x+xy)+1/(1+y+yz)+1/(xyz+z+xz)
=1/x(yz+1+y)+1/(1+y+yz)+1/z(xy+1+x)
=1/x(yz+1+y)+1/(1+y+yz)+1/z(xy+xyz+x)
=1/x(yz+1+y)+1/(1+y+yz)+1/xz(y+yz+1)
=(z+xz+1)/xz(y+yz+1)
=(z+xz+1)/(xyz+xyz*z+xz)
=(z+xz+1)/(1+z+xz)
=1