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求证：1+1/a+(a+1)(/ab+(a+1)(b+1)/abc+(a+1)(b+1)(c+1)/abcd=(a+1)(b+1)(c+1)(d+1)/abcd

分类: 作业答案 • 2021-12-30 12:39:02
问题描述：

求证：1+1/a+(a+1)(/ab+(a+1)(b+1)/abc+(a+1)(b+1)(c+1)/abcd=(a+1)(b+1)(c+1)(d+1)/abcd
1+1/a+(a+1)(/ab+(a+1)(b+1)/abc+(a+1)(b+1)(c+1)/abcd=(a+1)(b+1)(c+1)(d+1)/abcd.

答

左面通分,或者把右面的括号打开
(1)左面=abcd/abcd+bcd/abcd+(a+1)cd/abcd+(a+1)(b+1)d/abcd+(a+1)(b+1)(c+1)/abcd
=[abcd+bcd+(a+1)cd+(a+1)(b+1)d+(a+1)(b+1)(c+1)]/abcd
=[(a+1)bcd+(a+1)cd+(a+1)(b+1)d+(a+1)(b+1)(c+1)]/abcd
=[(a+1)cd(b+1)+(a+1)(b+1)d+(a+1)(b+1)(c+1)]/abcd
=[(a+1)(b+1)d(c+1)+(a+1)(b+1)(c+1)]/abcd
=右面
(2)右面=[(a+1)(b+1)(c+1)d+(a+1)(b+1)(c+1)]/abcd
=[(a+1)(b+1)c+(a+1)(b+1)]/abc+(a+1)(b+1)(c+1)/abcd
=[(a+1)b+(a+1)]/ab+(a+1)(b+1)/abc+(a+1)(b+1)(c+1)/abcd
=(a+1)/a+(a+1)/ab+(a+1)(b+1)/abc+(a+1)(b+1)(c+1)/abcd
=右面

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