设全集U={1,2,3,4,5,6,7,8,9}若A∩B={3},A∩(CuB)={1,5,7},(CuA)∩(CuB)
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设全集U={1,2,3,4,5,6,7,8,9}若A∩B={3},A∩(CuB)={1,5,7},(CuA)∩(CuB)={9},求A,B
要 的
要 的
![设全集U={1,2,3,4,5,6,7,8,9}若A∩B={3},A∩(CuB)={1,5,7},(CuA)∩(CuB)](/uploads/image/z/153808-16-8.jpg?t=%E8%AE%BE%E5%85%A8%E9%9B%86U%3D%7B1%2C2%2C3%2C4%2C5%2C6%2C7%2C8%2C9%7D%E8%8B%A5A%E2%88%A9B%3D%7B3%7D%2CA%E2%88%A9%28CuB%29%3D%7B1%2C5%2C7%7D%2C%28CuA%29%E2%88%A9%28CuB%29)
A={1,3,5,7}
B={2,3,4,6,8}
由A∩B={3}知A、B中都有3,初步猜测A={3……
B={3……
由A∩(CuB)={1,5,7},得A中有1 5 7
再猜A={1,3,5,7……} B={3……
CuB={1,5,7……}
由(CuA)∩(CuB)={9},知A B中都没9,即CuAB中都有9
则A={1,3,5,7……} B={3……}
因为CuA+A=U CuB+B=U
所以可猜测CuA={2,4,6,8,9}
CuB={1,5,7,9}(因为AB中157和246一定要错开,分别在原集和补集中)
验算下
A={1,3,5,7}
B={2,3,4,6,8}
CuA={2,4,6,8,9}
CuB={1,5,7,9}
说实话这东西没多大技术含量的,没办法给你很严谨的去证明
B={2,3,4,6,8}
由A∩B={3}知A、B中都有3,初步猜测A={3……
B={3……
由A∩(CuB)={1,5,7},得A中有1 5 7
再猜A={1,3,5,7……} B={3……
CuB={1,5,7……}
由(CuA)∩(CuB)={9},知A B中都没9,即CuAB中都有9
则A={1,3,5,7……} B={3……}
因为CuA+A=U CuB+B=U
所以可猜测CuA={2,4,6,8,9}
CuB={1,5,7,9}(因为AB中157和246一定要错开,分别在原集和补集中)
验算下
A={1,3,5,7}
B={2,3,4,6,8}
CuA={2,4,6,8,9}
CuB={1,5,7,9}
说实话这东西没多大技术含量的,没办法给你很严谨的去证明
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