求 1⼀a+（1⼀b)（1+1⼀a）+（1⼀c)(1+1⼀a)(1+1⼀b)+(1⼀d)(1+1⼀a)(1+1⼀b)(1+1⼀c)-(1+1⼀a)(1+1⼀b)(1+⼀c)(1+1⼀d)

1/a+(1/b)(1+1/a)+(1/c)(1+1/a)(1+1/b)+(1/d)(1+1/a)(1+1/b)(1+1/c)-(1+1/a)(1+1/b)(1+/c)(1+1/d)

=1/a+(1/b)[(a+1)/a]+(1/c)[(a+1)/a][(b+1)/b]+(1/d)[(a+1)/a][(b+1)/b][(c+1)/c]-[(a+1)/a][(b+1)/b][(c+1)/c][(d+1)/d]

=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)

=bcd/(abcd)+cd(a+1)/(abcd)+d(a+1)(b+1)/(abcd)+(a+1)(b+1)(c+1)/(abcd)-(a+1)(b+1)(c+1)(d+1)/(abcd)

=[bcd+cd(a+1)+d(a+1)(b+1)+(a+1)(b+1)(c+1)-(a+1)(b+1)(c+1)(d+1)]/(abcd)

={bcd+(a+1)[(cd+d(b+1)+(b+1)(c+1)-(b+1)(c+1)(d+1)]}/(abcd)

={bcd+(a+1){{cd+(b+1)[d+(c+1)-(c+1)(d+1)]}}}/(abcd)

={bcd+(a+1){{cd+(b+1){d+(c+1)[1-(d+1)]}}}}/(abcd)

={bcd+(a+1){{cd+(b+1)[d-d(c+1)]}}}/(abcd)

={bcd+(a+1){{cd+(b+1){d[1-(c+1)]}}}}/(abcd)

={bcd+(a+1)[cd-cd(b+1)]}/(abcd)

={bcd+(a+1){cd[1-(b+1)]}/(abcd)

=[bcd-bcd(a+1)]/(abcd)

={bcd[1-(a+1)]}/(abcd)

=(-abcd)/(abcd)

=-1