运用运算律计算:1/x+y+z*(1/x+1/y+1/z)×1/xy+yz+zx*1/xy+1/yz+1/zx

学习 时间:2026-04-02 17:24:21 阅读:4030
运用运算律计算:1/x+y+z*(1/x+1/y+1/z)×1/xy+yz+zx*1/xy+1/yz+1/zx

最佳回答

悲凉的长颈鹿

朴实的萝莉

2026-04-02 17:24:21

你的表达可能有点问题,是不是想求:[1/(x+y+z)](1/x+1/y+1/z)[1/(xy+yz+zx)][1/(xy)+1/(yz)+1/(zx)]?若是这样,则方法如下:∵1/x+1/y+1/z=(yz+zx+xy)/(xyz),∴(1/x+1/y+1/z)[1/(xy+yz+zx)]=1/(xyz)。∵1/(xy)+1/(yz)+1/(zx)=(z+x+y)/(xyz),∴[1/(x+y+z)])][1/(xy)+1/(yz)+1/(zx)]=1/(xyz)。∴原式=1/(xyz)^2。注:若原题不是我所猜测的那样,则请你补充说明。

最新回答共有2条回答

  • 壮观的西装
    回复
    2026-04-02 17:24:21

    你的表达可能有点问题,是不是想求:[1/(x+y+z)](1/x+1/y+1/z)[1/(xy+yz+zx)][1/(xy)+1/(yz)+1/(zx)]?若是这样,则方法如下:∵1/x+1/y+1/z=(yz+zx+xy)/(xyz),∴(1/x+1/y+1/z)[1/(xy+yz+zx)]=1/(xyz)。∵1/(xy)+1/(yz)+1/(zx)=(z+x+y)/(xyz),∴[1/(x+y+z)])][1/(xy)+1/(yz)+1/(zx)]=1/(xyz)。∴原式=1/(xyz)^2。注:若原题不是我所猜测的那样,则请你补充说明。

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