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

问题描述:

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∩CuB
y≠x+1,不是y≠x=1

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)}