设集合A={x|-3≤2x-1≤3},集合y=lg(x-1)定义域,则A∩B= A(1,2) B[-4,-6] C[-2,-2) D(1,2]
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![设集合A={x|-3≤2x-1≤3},集合y=lg(x-1)定义域,则A∩B= A(1,2) B[-4,-6] C[-2,-2) D(1,2]](/uploads/image/z/12043924-52-4.jpg?t=%E8%AE%BE%E9%9B%86%E5%90%88A%3D%EF%BD%9Bx%7C-3%E2%89%A42x-1%E2%89%A43%EF%BD%9D%2C%E9%9B%86%E5%90%88y%3Dlg%EF%BC%88x-1%EF%BC%89%E5%AE%9A%E4%B9%89%E5%9F%9F%2C%E5%88%99A%E2%88%A9B%3D+A%EF%BC%881%2C2%EF%BC%89+B%5B-4%2C-6%5D+C%5B-2%2C-2%EF%BC%89+D%EF%BC%881%2C2%5D)
设集合A={x|-3≤2x-1≤3},集合y=lg(x-1)定义域,则A∩B= A(1,2) B[-4,-6] C[-2,-2) D(1,2]
设集合A={x|-3≤2x-1≤3},集合y=lg(x-1)定义域,则A∩B= A(1,2) B[-4,-6] C[-2,-2) D(1,2]
设集合A={x|-3≤2x-1≤3},集合y=lg(x-1)定义域,则A∩B= A(1,2) B[-4,-6] C[-2,-2) D(1,2]
A={x|-3≤2x-1≤3}={x|-11,即B={x|x>1}
所以A交B={x|1
解
集合A={x|-3≤2x-1≤3}={x|-1≤x-1≤2}
x-1>0,解得x>1
所以集合B={x| x>1}
所以 A∩B=(1,2]
所以答案为 D