题目内容
设集合A={x|a≤x≤a+3},B={x|x<-1或x>5},分别求下列条件下实数a的值构成的集合.(1)A∩B=
;(2)
; (3)
.

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(1)
(2)
(3) 
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
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(1)由题意得
得
所以实数a的值构成的集合为{
} ……………4分
(2) 因为
所以
所以有
所以实数a的值构成的集合为{
} ……………10分
(3)由题意得
得
这样得实数a不存在,所以实数a的值构成的集合为
…………14分


所以实数a的值构成的集合为{

(2) 因为


所以有

所以实数a的值构成的集合为{

(3)由题意得


这样得实数a不存在,所以实数a的值构成的集合为


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