3.指出下列等式和命题是否成立,并说明理由:-|||-(1) cup B=(AB)cup B;-|||-(2) overline (A)B=Acup B:-|||-(3) overline (Acup B)cap C=overline (AB)C;-|||-(4) (AB)(overline (AB))=phi ;-|||-(5)若A⊆B,则 =AB;-|||-(6)若 =varnothing , 且C⊆A,则 =varnothing ;-|||-(7)若A⊆B,则A⊆B;-|||-(8)若B⊆A,则 cup B=A.

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

二、化简下列各式-|||-1、 (Acup B)(overline (A)cup B)(Acup overline (B))

(5)设 (A)=dfrac (1)(3), (B)=dfrac (1)(4), (Acup B)=dfrac (1)(2), 则 (overline (A)c

1.已知事件A,B,C,则以下结论错误的是 () .-|||-(a) overline (AB)=overline (A)cup overline (B) (b

随机事件 A,B 满足 (A)=P(B)=dfrac (1)(2), 且 (Acup B)=1, 则 (overline (A)cup overline (B)

1.2.指出下列关系中那些是正确的,那些是错误的,并说明理由-|||-(1) (Acup B)-C=Acup (B-C);-|||-(2) (Acup B)-A

设 A，B，C 为三个事件，指出下列各等式成立的条件.(1) ABC = A(2) A cup B cup C = A(3) A cup B = AB(4) A

10.证明下列事件的运算公式:()-|||-(1) =ABcup Aoverline (B); __-|||-(2) cup B=Acup overline (

10.证明下列事件的运算公式:-|||-(1) =ABcup Aoverline (B);-|||-__-|||-(2) cup B=Acup overline

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