x^2-2xy+2y^2+4y+4=0,求x^y
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x^2-2xy+2y^2+4y+4=0,求x^y
x^2-2xy+2y^2+4y+4=0,求x^y
x^2-2xy+2y^2+4y+4=0,求x^y
x^2-2xy+2y^2+4y+4=0
配方得:
(x-y)²+(y+2)²=0
所以
x-y=0(1)
y+2=0(2)
由(2)解得:
y=-2代入(1)
x+2=0
x=-2;
所以:
x^y
=(-2)^-2
=1/(-2)²
=1/4
-0.25
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