已知x2+y2+z2+14=2x-4y+6z,求x+y+z的值

Answer 1

x^2+y^2+z^2+14=2x-4y+6z
(x^2-2x+1)+(y^2+4y+4)+(z^2-6z+9)=0
(x-1)^+(y+2)^2+(z-3)^2=0
x=1、y=-2、z=3
x+y+z=2

Answer 2

x2+y2+z2+14=2x-4y+6z
(x-1)^2+(y+2)^2+(z-3)^2=0
x=1,y=-2,z=3
x+y+z = 2

Answer 3

（x-1）2+（y+2）2+（z-3）2=0
故：X=1，Y=-2，Z=3
X+Y+Z=六