2.1.2 用逻辑代数定律证明下列等式:-|||-(1) +overline (A)B=A+B-|||-__-|||-(2) +Aoverline (B)C+ABoverline (C)=AB+AC-|||-(3) +Aoverline (B)overline (C)+overline (A)CD+(overline (C)+overline (D))E=A+CD+E-|||-(4) (A+B)(A+overline (B))=A-|||-(5) +overline (A)C+BCD=AB+overline (A)C-|||-(6) (A+B)(B+C)(A+C)=(A+B)(A+C)

参考答案与解析:

相关试题

(1)(overline(AB)cup C)(overline(AC)); (2)(Acup B)(Acupoverline(B)).6.证明:(Acup B)-B=overline(AB)=A-A

(1)(overline(AB)cup C)(overline(AC)); (2)(Acup B)(Acupoverline(B)).6.证明:(Acup B

  • 查看答案
  • [题目]-|||-证明下列逻辑恒等式(方法不限)-|||-(1) AB'+B+A'B=A+B-|||-(2) (A+C')(B+D)(B+D')=AB+BC'-

    [题目]-|||-证明下列逻辑恒等式(方法不限)-|||-(1) AB+B+AB=A+B-|||-(2) (A+C)(B+D)(B+D)=AB+BC-|||-(

  • 查看答案
  • [题2.2] 证明下列逻辑恒等式(方法不限)-|||-(1) '+B+A'B=A+B-|||-(2) (A+C')(B+D)(B+D')=AB+BC'-|||-

    [题2.2] 证明下列逻辑恒等式(方法不限)-|||-(1) +B+AB=A+B-|||-(2) (A+C)(B+D)(B+D)=AB+BC-|||-(3) (

  • 查看答案
  • 2.3.1 用代数法将下列各式化简成最简的与-或表达式:-|||-(1) overline (AB+overline {A)B}+overline (A)B+Aoverline (B)-|||-(2)

    2.3.1 用代数法将下列各式化简成最简的与-或表达式:-|||-(1) overline (AB+overline {A)B}+overline (A)B+A

  • 查看答案
  • 1.下列等式不成立的是 () .-|||-(A) =ABcup Aoverline (B); (B) -B=Aoverline (B);-|||-(C) (overrightarrow (AB))(o

    1.下列等式不成立的是 () .-|||-(A) =ABcup Aoverline (B); (B) -B=Aoverline (B);-|||-(C) (ov

  • 查看答案
  • 5.化简:-|||-(1)(AB∪C)(AC);-|||-(2) (Acup B)(Acup overline (B)).

    5.化简:-|||-(1)(AB∪C)(AC);-|||-(2) (Acup B)(Acup overline (B)).

  • 查看答案
  • 三、判断以下等式是否成立1) A∪B=AB∪B; ( )2) overline(A∪BC)=overline(B∪C); ( )3) 若overline(A)subsetoverline(B)

    三、判断以下等式是否成立1) A∪B=AB∪B; ( )2) overline(A∪BC)=overline(B∪C); ( )3) 若overline

  • 查看答案
  • 2.设A,B为二随机事件,则 (ABcup Aoverline (B))(Acup overline (A)overline (B))= ()-|||-(A)A (B)B (C)AB (D)∅

    2.设A,B为二随机事件,则 (ABcup Aoverline (B))(Acup overline (A)overline (B))= ()-|||-(A)A

  • 查看答案
  • 2.设A,B为二随机事件,则 (ABcup Aoverline (B))(Acup overline (A)overline (B))= ()-|||-(A)A (B)B (C)AB (D) ∅

    2.设A,B为二随机事件,则 (ABcup Aoverline (B))(Acup overline (A)overline (B))= ()-|||-(A)A

  • 查看答案
  • 2.3.1 用代数法将下列各式化简成最简的与-或表达式:-|||-(1) overline (AB+overline {A)}overline (B)+overline (A)B+Aoverline

    2.3.1 用代数法将下列各式化简成最简的与-或表达式:-|||-(1) overline (AB+overline {A)}overline (B)+over

  • 查看答案