(5F(A, B, C, D, E, F)=AC+ABC+BC+ABCDEF A: F=AC+AbC+BC +ABCDEF =C(A+AB+B)+ABCDEF C(A+1)+ABCDEF =C+ABCDEF解: F= A C + A B C + B C + A B C D E F = C ( A + A B + B ) + A B C D E F = C ( A + 1 ) + A B C D E F = C + A B C D EF = C