y=x+1时代入(y-3)/(x-2)=1,(x+1-3)/(x-2)=1,(x-2)/(x-2)=1所以x为不等于2的任意实数,所以
Cu(A∪B)={(x,y)|x=2,y=3}
设全集U={(x,y)|x∈R,y∈R}集合A={(x,y)|(y-3)/(x-2)=1}B={(x,y)|y≠x+1}求:Cu(A∪B)
设全集U={(x,y)|x∈R,y∈R}集合A={(x,y)|(y-3)/(x-2)=1}B={(x,y)|y≠x+1}求:Cu(A∪B)
数学人气:534 ℃时间:2019-12-04 08:07:30
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