using boolean algebra transform the following expression into :sum of minterms form.F(A,B,C)=(A+B.C').(A.B+C')

Answer 1

F(A, B, C) = (A + BC) (AB + C)

First off, distribute it like you would a mathematical binomial.

(A + BC) (AB + C) = AAB + AC + BCAB + BC

Note that boolean algebra states two duplicate terms can merge into one. This then becomes

= AB + AC + ABC + BC

Swapping terms around,

= AB + ABC + AC + BC

And then factoring the first two terms

= AB (1 + C) + AC + BC

1 + C is equivalent to (true or C), and, in a disjunction, (true or C) is just "true"

= AB(1) + AC + BC
= AB + AC + BC

That should be in its reduced form.

Answer 2

i got here across an extra straightforward thank you to try this. (A+B)(A+B')(A'+C) = (A+BB')(A'+C), from distributive sources of "or" over "and" = (A+fake)(A'+C) = A(A'+C) = AA' + AC, from distributive sources of "and" over "or" = fake + AC = AC Lord bless you at the instant!

Answer 3

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

distribute the products:

=A(AB) + A(C') + BC'(AB) + BC'(C')

=AAB + AC' +ABBC' +BC'C'

use AA = A

=AB + AC' + ABC' + BC'

fill in (example AB = ABC + ABC')

=ABC + ABC' +ABC' + AB'C' + ABC' +ABC' + A'BC'

(ABC' + ABC' = ABC')

=ABC + ABC' +AB'C' +A'BC'

which is 4 out of 8 possibilities.