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

1个回答

  • 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