Варианты заданий
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25Вариант 8
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A∪B=A∩B,
C⊆D ⇒ A×C⊆B×D. -
(1− 1 )(1− 1 )(1− 1 ) … (1− 1 )= n+1 для n≥2. 4 9 16 n2 2n - |Z×Q|=ω.
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P1 = {〈a,1〉,〈a,4〉,〈b,2〉,〈b,3〉,〈c,1〉,〈c,4〉},
P2 = {〈1,1〉,〈1,4〉,〈2,1〉,〈3,4〉,〈4,3〉,〈4,1〉}. - P⊆R2, 〈x,y〉∈P⇔x−y∈Z.
- 〈R;√,−〉.
- B=〈Z;+,−〉, X={4,10}.
- G1: G2:
- G:
- (x↔y)∨(y↓x), ((x→y)|z)⊕xy.
- x|(y→z) и (x|y)→(x|z).
- (x∨y)→(z⊕x).
- f(0,1,1)=f(1,0,0)=f(1,1,0)=0.
- (1111 1100 1011 1011).
- J={x→y, x∧y}.
- (A∪B)\(C∩B)=(A\C)∪(A\B).