作业帮若x.y满足x^2+4x+y^2-10y+29=0.求x+y的值
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若x.y满足x^2+4x+y^2-10y+29=0.求x+y的值
若x.y满足x^2+4x+y^2-10y+29=0.求x+y的值
若x.y满足x^2+4x+y^2-10y+29=0.求x+y的值
x^2+4x+y^2-10y+29=0
把29拆成4和25
(x^2+4x+4)+(y^2-10y+25)=0
(x+2)^2+(y-5)^2=0
平方都是≥0的(x+2)^2≥0 (y-5)^2≥0
因为(x+2)^2+(y-5)^2=0
所以(x+2)^2=0 (y-5)^2=0
x=-2 y=5
所以x+y=5-2=3
x^2+4x+y^2-10y+29=0
x^2+4x+4+y^2-10y+25=0
(x+2)^2+(y-5)^2=0
(x+2)^>=0;(y-5)^2>=0
x=-2;y=5
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