若x^2+2x+y^2-6y+10=0,求x,y的值

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若x^2+2x+y^2-6y+10=0,求x,y的值
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若x^2+2x+y^2-6y+10=0,求x,y的值
若x^2+2x+y^2-6y+10=0,求x,y的值

若x^2+2x+y^2-6y+10=0,求x,y的值
x^2+2x+y^2-6y+10=0
(x^2+2x+1)+(y^2-6y+9)=0
(x+1)^2+(y-3)^2=0
∴x+1=0,y-3=0
∴x=-1,y=3

将方程配方,把10拆成1和9，则
x^2+2x+1+y^2-6y+9=0
（x+1）^2+(y-3)^2=0
由于平方数的非负性，
所以x+1=0，y-3=0
即x=-1，y=3

(x+1)^2+(y-3)^2=0
所以只能x+1=y-3=0
x=-1 y=3

原式=[（x+1）^2-1]+[(y-3)^2-9]+10=0
即：(x+1)^2+(y-3)^=0
所以他们分别等于0
得：x=-1,y=3
……学习快乐……

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