Question

已知x^2+y^2-6x+10y+34=0,求x+y的值

Answer 1

x^2+y^2-6x+10y+34=0

x^2-6x+9+y^2+10y+25=0

(x-3)^2+(y+5)^2=0

因为(x-3)^2>=0 (y+5)^2>=0

所以(x-3)^2=(y+5)^2=0时成立

x=3 y=-5

x+y=-2