求(1-二分之一-三分之一-四分之一-五分之一)(二分之一+三分之一+四分之一+五分之一+六分之一)-

设A=1/2+1/3+1/4+1/5+1/6,且B=1/2+1/3+1/4+1/5

则有原式=(1-B)A-(1-A)B

=A-AB-(B-AB)

=A-AB-B+AB

=A-B

将上述值代入此式

当A=1/2+1/3+1/4+1/5+1/6,B=1/2+1/3+1/4+1/5时

原式=A-B=1/6

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(1 - 1/2 - 1/3 - 1/4 - 1/5)(1/2 + 1/3 + 1/4 + 1/5 + 1/6) - (1 - 1/2 - 1/3 - 1/4 - 1/5 - 1/6)(1/2 + 1/3 + 1/4 + 1/5)
= [1 - (1/2 + 1/3 + 1/4 + 1/5)] * [(1/2 + 1/3 + 1/4 + 1/5) + 1/6] - [1 - (1/2 + 1/3 + 1/4 + 1/5) - 1/6] * (1/2 + 1/3 + 1/4 + 1/5)
避免眼睛混乱，假设 a = (1/2 + 1/3 + 1/4 + 1/5)

= (1 - a) * (a + 1/6) - (1 - a - 1/6) * a
= (1 - a) * a + (1-a) * (1/6) - (1-a) * a + (1/6) * a
= 1/6 - 1/6* a + 1/6 * a
= 1/6

是这题吗？(1-1/2-1/3-1/4-1/5)(1/2+1/3+1/4+1/5+1/6)-(1-1/2-1/3-1/4-1/5-1/6)(1/2+1/3+1/4+1/5)
解答：令a=1-1/2-1/3-...-1/6;b=1/2+1/3+...+1/6;
则原式=(a+1/6)×b-a×(b-1/6);
=ab+b/6-ab+a/6;
=(a+b)/6;
=1/6;