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R=1+2xy+(xy)^2, what is the value of x∗y ?
(1) R = 0
(2) x > 0
==> transforming the original condition and the question by variable approach method, we have, R=(1+xy)^2, 1+xy=R, -R, then xy=R-1, xy=-R-1. if we know R we can find xy
In case of 1), if R=0 then xy=-1, a unique answer. Therefore it is sufficient. Thus A is the answer.
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