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_数学解答

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

人气：281 ℃　时间：2019-08-19 19:21:56

解答

1.xyz=1x/(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)/...