Respuesta :

Answer:

A+A'C'

Explanation:

let F= AB+(AC)'+AB'C(AB+C)

       =AB +A'C'+AB'CAB+AB'CC

as we know that BB'=0 ( multiplicative inverse )

CC=C and AA=A

putting BB'=0 , CC=C and AA=A  in the expression of F we can find

F= AB +A'C'+0+AB'

F=AB+A'C'+AB'

F=AB+AB'+A'C'

taking A common

F=A(B+B')+A'C'

as B+B'=1 (additive inverse )

F=A+A'C'

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