已知x+y+z=1,x²+y²+z²=2求xy+yz+zx

已知x+y+z=1,x²+y²+z²=2求xy+yz+zx
已知x+y+z=1,x²+y²+z²=2求xy+yz+zx

已知x+y+z=1,x²+y²+z²=2求xy+yz+zx
(x+y+z)²=x²+y²+z²+2xy+2yz+2xz
所以可得：
xy+yz+xz
=[(x+y+z)²-(x²+y²+z²)]/2
=[1²-2]/2
=-1/2

(x+y+z)^2=x²+y²+z²+2(xy+yz+zx)=2+2(xy+yz+zx)=1
xy+yz+zx=-1/2