Is x=y?
1) x^4+y^4=2x^2y^2
2) x^4+y^4=0
==> In the original condition, there are 2 variables (x,y) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), for con 1), from x^4+y^4-2x^2y^2=0, (x^2-y^2)2=0, you get x^2=y^2, and from x=±y, yes and no coexists, hence it is not sufficient. For con 2), you only get x=y=0, hence yes, it is sufficient.
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1) x^4+y^4=2x^2y^2
2) x^4+y^4=0
==> In the original condition, there are 2 variables (x,y) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), for con 1), from x^4+y^4-2x^2y^2=0, (x^2-y^2)2=0, you get x^2=y^2, and from x=±y, yes and no coexists, hence it is not sufficient. For con 2), you only get x=y=0, hence yes, it is sufficient.
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