已知全集u={1,2,3,4,5,6,7,},A={2,4,5},B={1,3,5,7}求(CuA)U(Cu B) , (CuB)∩(CuB),Cu(A∩B)Cu(A∪B)
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![已知全集u={1,2,3,4,5,6,7,},A={2,4,5},B={1,3,5,7}求(CuA)U(Cu B) , (CuB)∩(CuB),Cu(A∩B)Cu(A∪B)](/uploads/image/z/3671224-16-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%85%A8%E9%9B%86u%3D%7B1%2C2%2C3%2C4%2C5%2C6%2C7%2C%7D%2CA%3D%7B2%2C4%2C5%7D%2CB%3D%7B1%2C3%2C5%2C7%7D%E6%B1%82%28CuA%29U%28Cu+B%29+%2C+%28CuB%29%E2%88%A9%EF%BC%88CuB%29%2CCu%EF%BC%88A%E2%88%A9B%29Cu%EF%BC%88A%E2%88%AAB%29)
已知全集u={1,2,3,4,5,6,7,},A={2,4,5},B={1,3,5,7}求(CuA)U(Cu B) , (CuB)∩(CuB),Cu(A∩B)Cu(A∪B)
已知全集u={1,2,3,4,5,6,7,},A={2,4,5},B={1,3,5,7}求(CuA)U(Cu B) , (CuB)∩(CuB),Cu(A∩B)
Cu(A∪B)
已知全集u={1,2,3,4,5,6,7,},A={2,4,5},B={1,3,5,7}求(CuA)U(Cu B) , (CuB)∩(CuB),Cu(A∩B)Cu(A∪B)
CuA=U-A={1,3,6,7} CuB=U-B={2,4,6}
(CuA)U(CuB)={1,2,3,4,6,7}
(CuA)∩(CuB)={6}
Cu(A∩B)=(CuA)U(CuB)={1,2,3,4,6,7}
Cu(AUB)=CuA∩CuB={6}
CuA={1,3,6,7} CuB={2,4,6} 所以(CuA)并(CuB)={1,2,3,4,6,7} (CuA)(题打错了吧)交(CuB)={6} A交B={5},所以Cu(A交B)={1,2,3,4,6,7} A并B={1,2,3,4,5,7},所以
Cu(A并B)={6}
已知全集U={1,2,3,4,5,6,7},A={2,4,5},b={1,3,5,7}求An(CuB)n(CuB).
(CuA)U(Cu B)={1,3,6,7}U{2,4,6}={1,2,3,4,6,7}
(CuB)∩(CuB)={2,4,6}
Cu(A∩B)={1,2,3,4,6,7,},
Cu(A∪B)={6}