求(1+1/2+1/2+1/4+1/5)*1-(1+1/2+1/3+1/4)的计算过程.
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![求(1+1/2+1/2+1/4+1/5)*1-(1+1/2+1/3+1/4)的计算过程.](/uploads/image/z/630443-11-3.jpg?t=%E6%B1%82%281%2B1%2F2%2B1%2F2%2B1%2F4%2B1%2F5%29%2A1-%281%2B1%2F2%2B1%2F3%2B1%2F4%29%E7%9A%84%E8%AE%A1%E7%AE%97%E8%BF%87%E7%A8%8B.)
求(1+1/2+1/2+1/4+1/5)*1-(1+1/2+1/3+1/4)的计算过程.
求(1+1/2+1/2+1/4+1/5)*1-(1+1/2+1/3+1/4)的计算过程.
求(1+1/2+1/2+1/4+1/5)*1-(1+1/2+1/3+1/4)的计算过程.
(1+1/2+1/3+1/4+1/5)*1-(1+1/2+1/3+1/4)
=1+1/2+1/3+1/4+1/5-1-1/2-1/3-1/4
=1/5
(1+1/2+1/2+1/4+1/5)*1-(1+1/2+1/3+1/4)
=1*1*1/5
=1/5
因为
1+1/2+1/3+1/4 重复出现,先加后减等于0