已知x2 + y2 + z2 = xy + xz + yz = 3 求x+y+z
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已知x2+y2+z2=xy+xz+yz=3求x+y+z已知x2+y2+z2=xy+xz+yz=3求x+y+z已知x2+y2+z2=xy+xz+yz=3求x+y+zx+y+z平方得x2+y2+z2+2x
已知x2 + y2 + z2 = xy + xz + yz = 3 求x+y+z
已知x2 + y2 + z2 = xy + xz + yz = 3 求x+y+z
已知x2 + y2 + z2 = xy + xz + yz = 3 求x+y+z
x+y+z平方得
x2+y2+z2+2xy+2xz+2yz吧
=9
所以x+y+z=3或者-3
（ps：x=y=z=1或者x=y=z=-1）
(x+y+z)2=x2+y2+z2+2xy+2yz+2xz=9
x+y+z=3或-3
(x+y+z)²=x²+y²+z²+2xy+2xz+2yz=(x²+y²+z²)+2(xy+xz+yz)=3+2*3=9
所以:x+y+z=3或-3
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