(x+1)^2+y^2+4-4y=0,则x+y=
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(x+1)^2+y^2+4-4y=0,则x+y=
(x+1)^2+y^2+4-4y=0,则x+y=
(x+1)^2+y^2+4-4y=0,则x+y=
(x+1)^2+y^2+4-4y=0
(x+1)²+(y-2)²=0
所以x+1=0,y-2=0
所以x=-1,y=2
x+y=-1+2=1
(x+1)^2+y^2+4-4y=(x+1)^2+(y-2)^2=0;
x=-1,y=2;
x+y=-1+2=1
等式化为(x+1)²+(y-2)²=0,x+1=0,y-2=0,x=-1,y=2.x+y=1
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