- 1
Simplify :(A+C). (A+A. D)+A. C+C
plz solve this question plzz anybody help
10 Respostas
+ 2
(a+c)(a+ad)+ac+c
= aa+aad+ac+acd+ac+c
= a(a+ad+2c+cd)+c
= a(a(1+d)+c(2+d))+c
but ... I do not really know what you are aiming for ...
+ 1
Thank u
+ 1
=AA+AAD+AC+ACD+AC+C
=AA(1+D)+AC(1+D+1)+C
=AA+AC+C
+ 1
Can you explain last and second step transition please ? It seems wrong to me @Shade96 :/
+ 1
=(AA+AAD)+(AC+ACD+AC)+C
=AA(1+D)+AC(1+D+1)+C
*I factor out (AA) it remain (1+D) and also factoring out AC from second parentheses
=AA+AC(2+D)+C
*logically (1+D)=1 (Boolean algebra theorem) it remain 1*AA which is equal AA
And about second term, yes I made a mistake thank you for warning me
@Baptiste
+ 1
He never said it was Boolean !
+ 1
So what is it -_-!
0
Numbers maybe :p
0
=AA+AAD+AC+ACD+AC+C
=AA(1+D)+AC(1+D+1)+C
=AA+AC+C
=A+C(A+1)
=A+C
LAST EXPRESSION
0
Ans. (A+ C) (A+AD) + AC + C
(A + C)A+ AC+ C
A+ AC+ AC + C
A+C