布尔代数化简 ((AB'+ABC)'+A(B+AB'))'

如题所述

第1个回答  2019-03-09
F=AB+A'C+B'C=AB+(A'+B')C=AB+(AB)'C=AB+C
F=(A+B')(B+C')(C+D')(D+A')
=A(B+C')(C+D')(D+A')+B'(B+C')(C+D')(D+A')
=(AB+AC')(C+D')(D+A')+B'C'(C+D')(D+A')
=AB(C+D')(D+A')+B'C'(C+D')(D+A')
=ABC(D+A')+B'C'D'(D+A')
=ABCD+A'B'C'D本回答被网友采纳
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