3.13计算 cup B cap B -B oplus B.-|||-(1) = a,b ,c} = c,d .-|||-(2) = a,{ b , c,(c),(a,b)}, = a,b ,c, b .-|||-(3) = x|xin Ncap xlt 3 = x|xin Ncap xgeqslant 2 .-|||-(4) = x|xin Rcap xlt 1 = x|xin Zcap xlt 1 .-|||-(5) = x|xin Zcap xlt 0 = x|xin Zcap xgeqslant 2 .

1.证明:-|||-(1) (Acup B)|C=A|(Bcup C) ;-|||-(2) (Acup B)|C=(A|C)cup (B|C) -

A，B，C是任意事件，在下列各式中，不成立的是（）(A) (A-B)cup B=Acup B.-|||-(B) (Acup B)-A=B.-|||-(C) (A

[题目]证明 cup (Bcap C)=(Acup B)cap (Acup C)

[题目]-|||-1 1 1 1-|||-a b c d-|||-a^2 b^2 c^2 d^2 =(a-b)(a-c)(a -d)(b-c)(b-d)(c-|

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

((a+b-c))^2+(b+c-a)(c+a-b)-|||-(a+b-c),-|||-面 (a+a-b)(a+b-c)+b(a+b-c)(b+c-a)+c(b

[单选题]有如下程序： a=1:b=2:c=3 a=a+b:b=b+c:c=b+a If a＜＞3 Or b＜＞3 Then a=b-a:b=C-a:C=b+a End If Print a+b+c 运行后，输出的结果是( )。A．16B．3C．6D．8

b=1 (B) =2, b=-1 (C) =-2. b=1 (D) =-2, b=-1

= a,b,c,d . 由表-|||-a b c d-|||-a a b c d-|||-·b b d a c-|||-c c a b d-|||-d d c

计算-|||-(1) (acdot b)c-(acdot c)b;