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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

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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∪B=______.
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
∵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}.