In B, note that the denominator (1-x) cannot be equal to 0. When (1-x) is not equal to 0 (that is, x is not equal to 1), b also represents the straight line x+y= 1. Therefore, set b includes "all other points on the straight line x+y= 1, but does not include (1, 0)".
So CuB includes "all points not on the straight line x+y= 1, but including (1, 0)".
So (CuB)∩A={( 1, 0)}.
I hope I can help you. If you have any questions, please ask. If you are satisfied, please adopt them. Thank you. I wish you progress in your study!