【设xyz=1,求x/(xy+x+1)+y/(yz+y+1)-查字典问答网
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  【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】

  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

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2020-02-0700:43
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柯铁军

  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)/...

2020-02-07 00:44:56
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