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  • 求證: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

    求證: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.
    數(shù)學(xué)人氣:465 ℃時(shí)間:2020-03-28 14:46:29
    優(yōu)質(zhì)解答
    左面通分,或者把右面的括號(hào)打開(kāi)
    (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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