已知X²+y²+z²—2x+4x-6z+14=0,求x+y+z的值.

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已知X²+y²+z²—2x+4x-6z+14=0,求x+y+z的值.
已知X²+y²+z²—2x+4x-6z+14=0,求x+y+z的值.

已知X²+y²+z²—2x+4x-6z+14=0,求x+y+z的值.
X²+y²+z²—2x+4y-6z+14
=(x^2-2x+1)+(y^2+4y+4)+(z^2-6z+9)
=(x-1)^2+(y+2)^2+(z-3)^2
=0
x=1,y=-2,z=3
x+y+z=1-2+3=2

有没有打错??? 打错哦了

应该是这样的吧:
X²+y²+z²—2x+4y-6z+14=0
(x^2-2x+1)+(y^2+4y+4)+(z^2-6z+9)=0
(x-1)^2+(y+2)^2+(z-3)^2=0
x-1=0
y+2=0
z-3=0
x=1,y=-2,z=3
x+y+z=2

(x-1)²+(y+2)²+(z-3)²+0
x+y+z=2

董事长错题也能算,厉害!

原 式可化为:x^2-2x+1+y^2+4y+4+z^2-6z+9=0
即:(x-1)^2+(y+2)^2+(z-3)^2=0
故只能:(x-1)^2=0,(y+2)^2=0,(z-3)^2=0
即x=1,y=-2,z=3则x+y+z=1-2+3=2