Franklin Davey
07/08/2024 · High School
\( F=(A+\bar{B}) \cdot(\bar{A}+B) \cdot(\bar{A}+\bar{B}) \) implementar el circuito sin simplificar
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To simplify the given expression \( F=(A+\bar{B}) \cdot(\bar{A}+B) \cdot(\bar{A}+\bar{B}) \), we can use the distributive property and the complement property of Boolean algebra.
First, let's expand the expression using the distributive property:
\( F = (A\bar{A} + A\bar{B} + \bar{A}B + \bar{B}B) \cdot (\bar{A} + \bar{B}) \)
Now, simplify the terms with complements:
\( F = (0 + A\bar{B} + \bar{A}B + 0) \cdot (\bar{A} + \bar{B}) \)
\( F = (A\bar{B} + \bar{A}B) \cdot (\bar{A} + \bar{B}) \)
Now, apply the distributive property again:
\( F = A\bar{B}\bar{A} + A\bar{B}\bar{B} + \bar{A}B\bar{A} + \bar{A}B\bar{B} \)
Simplify the terms with complements:
\( F = 0 + A\bar{B} + 0 + 0 \)
\( F = A\bar{B} \)
Therefore, the simplified expression for \( F=(A+\bar{B}) \cdot(\bar{A}+B) \cdot(\bar{A}+\bar{B}) \) is \( F = A\bar{B} \).
Quick Answer
\( F = A\bar{B} \)
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