已知x+y+z=1,x²+y²+z²=2求xy+yz+xz的值
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![已知x+y+z=1,x²+y²+z²=2求xy+yz+xz的值](/uploads/image/z/2515995-27-5.jpg?t=%E5%B7%B2%E7%9F%A5x%2By%2Bz%3D1%2Cx%26%23178%3B%2By%26%23178%3B%2Bz%26%23178%3B%3D2%E6%B1%82xy%2Byz%2Bxz%E7%9A%84%E5%80%BC)
已知x+y+z=1,x²+y²+z²=2求xy+yz+xz的值
已知x+y+z=1,x²+y²+z²=2求xy+yz+xz的值
已知x+y+z=1,x²+y²+z²=2求xy+yz+xz的值
(x+y+z)²=1 ,x²+2xy+y²+2(x+y)z+z²=1,x²+y²+z²+2(x+y)z+2xy=1
xy+yz+xz=-1/2
(x+y+z)×(x+y+z)=x²+y²+z²+2xy+2yz+2xz
(x+y)^=x^+y^+2xy=(1-z)^
(x+z)^=x^+z^+2xy=(1-y)^
(y+z)^=y^+z^+2yz=(1-x)^
相加得:
2(x^+y^+z^)+2(xy+yz+xz)=1-2z+z^+1-2y+y^+1-2x+x^
化简得:
4+2(xy+yz+xz)=3-2(x+y+z)+(x^+y^+z^)
所以xy+yz+xz=-0.5
希望能帮到你