已知集合A={x丨log1/2(3-x)>-2}B={X丨5/x+2≥1}则A∩B

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已知集合A={x丨log1/2(3-x)>-2}B={X丨5/x+2≥1}则A∩B
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已知集合A={x丨log1/2(3-x)>-2}B={X丨5/x+2≥1}则A∩B
已知集合A={x丨log1/2(3-x)>-2}B={X丨5/x+2≥1}则A∩B

已知集合A={x丨log1/2(3-x)>-2}B={X丨5/x+2≥1}则A∩B
解log1/2(3-x)>-2=log1/2(1/2)^(-2)=log1/2(4)
即log1/2(3-x)>log1/2(4)
即0<3-x<4
即-1<x<3
A={x丨-1<x<3}
5/x+2≥1
即5/x+2-1≥0
即5/x+2-(x+2)/(x+2)≥0
即[5-(x+2)]/(x+2)≥0
即(3-x)/(x+2)≥0
即(x-3)/(x+2)≤0
即(x-3)(x+2)≤0且x+2≠0
即-2<x≤3
即B={X丨-2<x≤3}
A∩B={x丨-1<x<3}