已知集合A={x丨x²+ax+12b=0}和B={x丨x²-ax+b=0}满足(CRA)∩B={2},A∩(CRB)={4}求实数a,b
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![已知集合A={x丨x²+ax+12b=0}和B={x丨x²-ax+b=0}满足(CRA)∩B={2},A∩(CRB)={4}求实数a,b](/uploads/image/z/1658163-3-3.jpg?t=%E5%B7%B2%E7%9F%A5%E9%9B%86%E5%90%88A%3D%EF%BD%9Bx%E4%B8%A8x%26%23178%3B%2Bax%2B12b%3D0%EF%BD%9D%E5%92%8CB%3D%EF%BD%9Bx%E4%B8%A8x%26%23178%3B-ax%2Bb%3D0%EF%BD%9D%E6%BB%A1%E8%B6%B3%28CRA%29%E2%88%A9B%3D%7B2%7D%2CA%E2%88%A9%28CRB%29%3D%7B4%7D%E6%B1%82%E5%AE%9E%E6%95%B0a%2Cb)
已知集合A={x丨x²+ax+12b=0}和B={x丨x²-ax+b=0}满足(CRA)∩B={2},A∩(CRB)={4}求实数a,b
已知集合A={x丨x²+ax+12b=0}和B={x丨x²-ax+b=0}满足(CRA)∩B={2},A∩(CRB)={4}求实数a,b
已知集合A={x丨x²+ax+12b=0}和B={x丨x²-ax+b=0}满足(CRA)∩B={2},A∩(CRB)={4}求实数a,b
(CRA)∩B={2}
即2属于B
同理4属于A
代入
16+4a+12b=0
4-2a+b=0
所以b=-12/7
a=8/7