已知集合A={(x,y)/2x—y=0},B={(x,y)/3x+y=0 }.C={(x,Y)/2x-y=3},求A∩B,A∩C (A∩B)∪(B∩C
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![已知集合A={(x,y)/2x—y=0},B={(x,y)/3x+y=0 }.C={(x,Y)/2x-y=3},求A∩B,A∩C (A∩B)∪(B∩C](/uploads/image/z/2713185-9-5.jpg?t=%E5%B7%B2%E7%9F%A5%E9%9B%86%E5%90%88A%3D%7B%28x%2Cy%29%2F2x%E2%80%94y%3D0%7D%2CB%3D%7B%EF%BC%88x%2Cy%EF%BC%89%2F3x%2By%3D0+%7D.C%3D%7B%EF%BC%88x%2CY%EF%BC%89%2F2x-y%3D3%7D%2C%E6%B1%82A%E2%88%A9B%2CA%E2%88%A9C+%EF%BC%88A%E2%88%A9B%EF%BC%89%E2%88%AA%EF%BC%88B%E2%88%A9C)
已知集合A={(x,y)/2x—y=0},B={(x,y)/3x+y=0 }.C={(x,Y)/2x-y=3},求A∩B,A∩C (A∩B)∪(B∩C
已知集合A={(x,y)/2x—y=0},B={(x,y)/3x+y=0 }.C={(x,Y)/2x-y=3},求A∩B,A∩C (A∩B)∪(B∩C
已知集合A={(x,y)/2x—y=0},B={(x,y)/3x+y=0 }.C={(x,Y)/2x-y=3},求A∩B,A∩C (A∩B)∪(B∩C
A∩B就是方程组2x—y=0,3x+y=0的解(0,0),A∩C就是方程组2x—y=0,2x-y=3的解空集,
同理(A∩B)∪(B∩C)就是两个方程组的并了,{(0,0),(0.6,-1.8)} .
A∩B={(0,0)},A∩C=空集,(A∩B)∪(B∩C)={(0,0),(0.6,-1.8)} 、题目时做出来了
但貌似我觉得是错了