(1/2+1/3+...+1/2000)*(1+1/2+...+1/1999)-(1+1/2+...+1/2000)*(1/2+1/3+...+1/1999)

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(1/2+1/3+...+1/2000)*(1+1/2+...+1/1999)-(1+1/2+...+1/2000)*(1/2+1/3+...+1/1999)
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(1/2+1/3+...+1/2000)*(1+1/2+...+1/1999)-(1+1/2+...+1/2000)*(1/2+1/3+...+1/1999)
(1/2+1/3+...+1/2000)*(1+1/2+...+1/1999)-(1+1/2+...+1/2000)*(1/2+1/3+...+1/1999)

(1/2+1/3+...+1/2000)*(1+1/2+...+1/1999)-(1+1/2+...+1/2000)*(1/2+1/3+...+1/1999)
分析:算式的第一部分,先用乘法分配律展开,就容易做了:
原式=(1/2+1/3+...+1/2000)+
(1/2+1/3+...+1/2000)×(1/2+...+1/1999)-(1+1/2+...+1/2000)*(1/2+1/3+...+1/1999)
=(1/2+1/3+...+1/2000)+(1/2+...+1/1999)×[(1/2+1/3+...+1/2000)-(1+1/2+...+1/2000)]
=(1/2+1/3+...+1/2000)+(1/2+...+1/1999)×(-1)
=(1/2+1/3+...+1/2000)+(-1/2-...-1/1999)
=1/2+1/3+…+1/2000-1/2-1/3-…-1/1999
=(1/2-1/2)+(1/3-1/3)+…+(1/1999-1/1999)+1/2000
=0+0+…+0+1/2000
=1/2000

设1&#47;2+1&#47;3+...+1&#47;1998 = a6则:(1&#47;2+1&#47;3+...+1&#47;1999)*(1+1&#47;2+...1&#47;1998)-(1+1&#47;2+...+1&#47;1999)*(1&#47;2+1&#47;3+...+1&#47;1998)=(a+ 1&#47;1999)*(1+a)-(1+a+1&#47;1999)*a= a...

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设1&#47;2+1&#47;3+...+1&#47;1998 = a6则:(1&#47;2+1&#47;3+...+1&#47;1999)*(1+1&#47;2+...1&#47;1998)-(1+1&#47;2+...+1&#47;1999)*(1&#47;2+1&#47;3+...+1&#47;1998)=(a+ 1&#47;1999)*(1+a)-(1+a+1&#47;1999)*a= a +1&#47;1999 + a&#178; +a&#47;1999 - a -a&#178; -a&#47;1999= 1&#47;1999

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