已知x²+y²+z²+2x-4y+6z+14=0 求x+y+z的值
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已知x²+y²+z²+2x-4y+6z+14=0 求x+y+z的值
已知x²+y²+z²+2x-4y+6z+14=0 求x+y+z的值
已知x²+y²+z²+2x-4y+6z+14=0 求x+y+z的值
x²+y²+z²+2x-4y+6z+14=0
x²+y²+z²+2x-4y+6z+1+4+9=0
x^2+2x+1+y^2-4y+4+z^2+6z+9=0
(x+1)^2+(y-2)^2+(z+3)^2=0
所以 x=-1,y=2,z=-3
x+y+z=-1+2-3=-2
原式 可化为 (x^2-2x+1)+(y^2+4y+4)+(z^2-6z+9)=0
(x-1)^2+(y+2)^2+(z-3)^2=0
所以 X=1 Y=-2 Z=3
所以X+Y+Z=1-2+3=2