,已知集合A={X||x—a|=4},集合B={1,2,b},（1）是否存在实数a的值,使得对于任意实数b都有A含于B?若存,求出对应的a,若不存在,试说明理由（2）若A含于B成立,求出对应的实数对（a,b）.

,已知集合A={X||x—a|=4},集合B={1,2,b},（1）是否存在实数a的值,使得对于任意实数b都有A含于B?若存,求出对应的a,若不存在,试说明理由（2）若A含于B成立,求出对应的实数对（a,b）.
,已知集合A={X||x—a|=4},集合B={1,2,b},（1）是否存在实数a的值,使得对于任意实数b都有A含于B?若存
,求出对应的a,若不存在,试说明理由（2）若A含于B成立,求出对应的实数对（a,b）.

,已知集合A={X||x—a|=4},集合B={1,2,b},（1）是否存在实数a的值,使得对于任意实数b都有A含于B?若存,求出对应的a,若不存在,试说明理由（2）若A含于B成立,求出对应的实数对（a,b）.
已知集合A={X||x—a|=4},集合B={1,2,b},（1）是否存在实数a的值,使得对于任意实数b都有A含于B?若存求出对应的a,若不存在,试说明理由（2）若A含于B成立,求出对应的实数对（a,b）.
(1)A={a+4,a-4},A中就两个元素.要不管b是多少,都有a含于B,则1,2就必须包括A中的所有元素,所以a+4=2,a-4=1,上面两个式子都成立,是解不出来的,所以不存在实数a,使得对于任意实数b都有a含于B.
（2）因为A含于B,所以a+4,a-4在B中都可以找到.现在就a+4等于B中的哪个数分类讨论.
①a+4=1,a-4=b得出a=-3,b=-7
②a+4=2,a-4=b得出a=-2,b=-6
③a+4=b,a-4=1得出a=5,b=9
a+4=b,a-4=2得出a=6,b=10
综合上述分析得出对应的实数对为
（-3,-7）、（-2,-6）、（5,9）、（6,10）

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Drop Ship Wholesalers
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Wholesale Dropship Companies - How you can Spot a Scam,wholesale ray ban sunglasses
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Drop Ship Wholesalers
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（1）由|x—a|=4，得x=a+4或a-4
A、令a+4=1得a=-3，则a-4=-7（不合题意）
B、令a+4=2得a=-2则a-4=-6（不合题意） 所以不存在
（2）由|x—a|=4，得x=a+4或a-4
A、令a+4=1得a=-3，由此得a-4=-7=b 所以（a，b）=（-3，-7）
B、令a+4=2得a=-2则a-4=-6=b 所以（a...

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（1）由|x—a|=4，得x=a+4或a-4
A、令a+4=1得a=-3，则a-4=-7（不合题意）
B、令a+4=2得a=-2则a-4=-6（不合题意） 所以不存在
（2）由|x—a|=4，得x=a+4或a-4
A、令a+4=1得a=-3，由此得a-4=-7=b 所以（a，b）=（-3，-7）
B、令a+4=2得a=-2则a-4=-6=b 所以（a，b）=（-2，-6）
C、令a-4=1得a=5则a+4=9=b 所以（a，b）=（5,9）
D、令a-4=2得a=6则a+4=10=b所以（a，b)=(6,10)
综上所述，实数对（a，b）=（-3，-7）或（-2，-6）或（5，9）或（6，10）

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