已知x2 + y2 + z2 = xy + xz + yz = 3 求x+y+z

问题描述:

已知x2 + y2 + z2 = xy + xz + yz = 3 求x+y+z

(x+y+z)&sup2=x&sup2+y&sup2+z&sup2+2xy+2xz+2yz=(x&sup2+y&sup2+z&sup2)+2(xy+xz+yz)=3+2*3=9
所以:x+y+z=3或-3

(x+y+z)2=x2+y2+z2+2xy+2yz+2xz=9
x+y+z=3或-3

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)