已知集合A={1,4,a2-2a},B={a-2,a2-4a+2,a2-3a+3,a2-5a},A∩B={1,3},则A

1个回答

  • 解题思路:由A∩B={1,3}得到a2-2a=3,解得:a=-1或a=3.然后分a=-1或a=3讨论,求出B,则A∪B可求.

    ∵A={1,4,a2-2a},B={a-2,a2-4a+2,a2-3a+3,a2-5a},且A∩B={1,3},

    ∴a2-2a=3,解得:a=-1或a=3.

    当a=-1时,a-2=-3,a2-4a+2=7,a2-3a+3=7,a2-5a=6.

    集合B违背集合中元素的互异性;

    当a=3时,a-2=1,a2-4a+2=-1,a2-3a+3=3,a2-5a=-6.

    B={1,-1,3,-6}.

    A∪B={1,-1,3,4,-6}.

    故答案为:{1,-1,3,4,-6}.

    点评:

    本题考点: 交集及其运算;并集及其运算.

    考点点评: 本题考查了交集、并集的运算,考查了集合中元素的特性,是基础题.