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.
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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.
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