若x^2+2x+y^2-6y+10=0,求x.y的值过程.没过程不采纳

2025-06-26 20:14:33
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回答1:

解:∵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。