问题描述：
设xyz=1,求x/（xy+x+1）+y/（yz+y+1）+z/（zx+z+1）的值
已知实数a满足a^2+4a-8=0,求1/(a+1)-(a+3)/(a^2-1)×(a^2-2a+1)/(a^2+6a+9)的值
问题解答：
我来补答
1.
xyz=1
x/(xy+x+1)+y/(yz+y+1)+z/(zx+z+1)将x/(xy+x+1)中的1换为xyz得:
=x/(xy+x+xyz)+y/(yz+y+1)+z/(zx+z+1)
=1/(yz+y+1)+y/(yz+y+1)+z/(zx+z+1)
=(1+y)/(yz+y+1)+z/(zx+z+1)将(1+y)/(yz+y+1)中的1换为xyz得:
=(xyz+y)/(yz+y+xyz)+z/(zx+z+1)
=(xz+1)/(zx+z+1)+z/(zx+z+1)
=(zx+z+1)/(zx+z+1)
=1
2.
a^2+4a-8=0
所以a^2+4a=8
1/(a+1)-(a+3)/(a^2-1)×(a^2-2a+1)/(a^2+6a+9)
=1/(a+1)-(a+3)/(a+1)(a-1)×(a-1)^2/(a+3)^2
=1/(a+1)-(a-1)/(a+1)(a+3)
=[(a+3)-(a-1)]/(a+1)(a+3)
=4/(a+1)(a+3)
=4/(a^2+4a+3)
=4/(8+3)
=4/11
