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Mathematics senior one u
In this problem, set A contains "all points on a straight line x+y =1";

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)}.

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