Question

数学代数题 求解答过程

已知X²+y²+z²—2x+4x-6z+14=0，求x+y+z的值。

Answer 1

X²前颤+y²慧纳败+z²茄碧—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

Answer 2

有没有打错？？？ 打错哦了

Answer 3

应该唤洞物是这和液颤局样的吧:
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

Answer 4

(x-1)²+(y+2)²+(z-3)²+0
x+y+z=2

Answer 5

董事长错题也能算，厉害！

Answer 6

解：原 式可知世卜化为返尘:x^2-2x+1+y^2+4y+4+z^2-6z+9=0
即搭穗:(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