用公式法简化逻辑函数为最简与或式:L=(A+B+C)(A+B+C')

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摘要 [F=A+ABC+A B ⋅ C +CB+ C ⋅ B =A(1+BC)+(A+1) B ⋅ C +CB=A+ B ⋅ C +CB]（2）[ F = A B ‾ C D + A B C ‾ ⋅ D ‾ + A B ‾ + A D ‾ + A B ‾ C = A B ‾ ( C D + 1 + C ) + A B C ‾ ⋅ D ‾ + A D ‾ = A B ‾ + A B C ‾ ⋅ D ‾ + A D ‾ = A B ‾ + A D ‾ ( B C ‾ + 1 ) = A ( B ‾ + D ‾ ) ] [F = A\overline{B}CD + AB\overline{C}·\overline{D} + A\overline{B} + A\overline{D} + A\overline{B}C = A\overline{B}(CD + 1 + C

咨询记录 · 回答于2022-10-15

用公式法简化逻辑函数为最简与或式:L=(A+B+C)(A+B+C')

[F=A+ABC+A B ⋅ C +CB+ C ⋅ B =A(1+BC)+(A+1) B ⋅ C +CB=A+ B ⋅ C +CB]（2）[ F = A B ‾ C D + A B C ‾ ⋅ D ‾ + A B ‾ + A D ‾ + A B ‾ C = A B ‾ ( C D + 1 + C ) + A B C ‾ ⋅ D ‾ + A D ‾ = A B ‾ + A B C ‾ ⋅ D ‾ + A D ‾ = A B ‾ + A D ‾ ( B C ‾ + 1 ) = A ( B ‾ + D ‾ ) ] [F = A\overline{B}CD + AB\overline{C}·\overline{D} + A\overline{B} + A\overline{D} + A\overline{B}C = A\overline{B}(CD + 1 + C

最终等于什么

是这样：Y=(A+B+C)(A'+B'+C')=AA'+BB'+CC'+AB'+BC'+CA'=AB'+BC'+CA'这是最简与或形式,用卡诺图可得到相同结果.

:L=(A+B+C)(A+B+C) C上面有一横的

ab+a+b

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