63)F(A,B,C=1A⊕BC=∑m(2 解:F(A,B,C)=ABC =A·BC+A·BC AB+C)+ABC ab+aC +ABC C A00011110 011 F(A,B,C=∑(0,1,2,7)(3)F(A,B,C)=1⊕A⊕BC=∑m(?) 解: F(A,B,C)= A ⊕ BC = A· BC + A ·BC =A (B + C ) +ABC = A B +A C +ABC F(A,B,C)= ∑( 0, 1, 2 ,7 )