若|x-y|+(y+3)2=0,则x2+y2-2xy=

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若|x-y|+(y+3)2=0,则x2+y2-2xy=若|x-y|+(y+3)2=0,则x2+y2-2xy=若|x-y|+(y+3)2=0,则x2+y2-2xy=∵|x-y|+(y+3)2=0∴x-y

若|x-y|+(y+3)2=0,则x2+y2-2xy=
若|x-y|+(y+3)2=0,则x2+y2-2xy=

若|x-y|+(y+3)2=0,则x2+y2-2xy=
∵|x-y|+(y+3)2=0
∴x-y=0
(y+3)2=0
∴x=-3 y=-3
∴x2+y2-2xy
=(x-y)²
=[-3-(-3)]²
=0

|x-y|+(y+3)^2=0,
则一定有 x-y=0
于是 x^2+y^2-2xy=(x-y)^2=0

∵|x-y|+(y+3)2=0
∴x-y=0
y+3=0
x2+y2-2xy=(x-y)²=0