已知全集U={(x,y)|x∈R,y∈R},A={(x,y)|(y-3)/(x-2)=1,x∈R,y∈R},B={(x,y)|y≠x=1,x∈R,y∈R},求CuA∩CuBy≠x+1,不是y≠x=1
网友回答
A = {(x,y)|(y-3)/(x-2)=1,x∈R,y∈R}
= {(x,y)|y=x+1,x≠2}
B = {(x,y)|y≠x+1},
则 A∪B = {(x,y)|y=x+1,x≠2 或 y≠x+1}
= {(x,y)|(x,y)≠(2,3)}
∴ CuA∩CuB
= Cu(A∪B)
= {(2,3)}