设A={(x,y)|2x-y=1},B={(x,y)|5x+y=6},C={(x,y)|2x=y+1}(1)A∩B,(2)C∩D,(3)A∩(CUD)D={(x,y)|2x-y=8
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![设A={(x,y)|2x-y=1},B={(x,y)|5x+y=6},C={(x,y)|2x=y+1}(1)A∩B,(2)C∩D,(3)A∩(CUD)D={(x,y)|2x-y=8](/uploads/image/z/11506882-58-2.jpg?t=%E8%AE%BEA%3D%7B%EF%BC%88x%2Cy%EF%BC%89%EF%BD%9C2x-y%3D1%7D%2CB%3D%7B%EF%BC%88x%2Cy%EF%BC%89%EF%BD%9C5x%2By%3D6%7D%2CC%3D%7B%EF%BC%88x%2Cy%EF%BC%89%EF%BD%9C2x%3Dy%2B1%7D%EF%BC%881%EF%BC%89A%E2%88%A9B%2C%EF%BC%882%EF%BC%89C%E2%88%A9D%2C%EF%BC%883%EF%BC%89A%E2%88%A9%EF%BC%88CUD%EF%BC%89D%3D%7B%EF%BC%88x%EF%BC%8Cy%EF%BC%89%EF%BD%9C2x-y%3D8)
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设A={(x,y)|2x-y=1},B={(x,y)|5x+y=6},C={(x,y)|2x=y+1}(1)A∩B,(2)C∩D,(3)A∩(CUD)D={(x,y)|2x-y=8
设A={(x,y)|2x-y=1},B={(x,y)|5x+y=6},C={(x,y)|2x=y+1}
(1)A∩B,(2)C∩D,(3)A∩(CUD)
D={(x,y)|2x-y=8
设A={(x,y)|2x-y=1},B={(x,y)|5x+y=6},C={(x,y)|2x=y+1}(1)A∩B,(2)C∩D,(3)A∩(CUD)D={(x,y)|2x-y=8
1.A∩B的结果即是2x-y=1和5x+y=6联立解出的结果:
两式相加得:7x=7,x=1,y=1
即A∩B={(1,1)}
2.同理
联立方程2x=y+1和2x-y=8解得:C∩D=空集
3.将原式A∩(CUD)化成*∩* 的形式后即可解出答案.
A∩(CUD)=A(或C)
D的集合呢?