(2F(A, B, C)=A+B C+AB +B+C 解:F=A+B·C+AB+B+C abc+AB+B+C =A(BC+B)+B+C A (B+C)+B+C =A+B+C+B +c =A+1解: F = A + B • C + A B + B + C = A B C + A B + B + C = A (B C + B) + B + C = A ( B + C ) + B + C = A + B + C + B + C = A + 1 = 1