若x^2-4x+y^2+2y+5=0,求x,y的值
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若x^2-4x+y^2+2y+5=0,求x,y的值若x^2-4x+y^2+2y+5=0,求x,y的值若x^2-4x+y^2+2y+5=0,求x,y的值x^2-4x+y^2+2y+5=0推出x^2-4x
若x^2-4x+y^2+2y+5=0,求x,y的值
若x^2-4x+y^2+2y+5=0,求x,y的值
若x^2-4x+y^2+2y+5=0,求x,y的值
x^2-4x+y^2+2y+5=0
推出x^2-4x+y^2+2y+4+1=0
(x-2)^2+(y+1)^2=0
因为(x-2)^2>=0 +(y+1)^2>=0
所以只能(x-2)^2=0 (y+1)^2=0 即x-2=0 y+1=0
所以x=2且 y=-1
X^2-4X+4+Y^2+2Y+1=0
(X-2)+(Y+1)=0
∴X=2 且 Y=负1
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