Question

已知x-y=2分之1+根号3,z-y=2分之1-根号3,求x²+y²+z²-xy-yz-xz的值

Answer 1

x^2+y^2+z^2-xy-yz-xz
=(1/2)[(x-y)^2+(z-y)^2+(z-x)^2]
x-y=1/2+根号3
z-y=1/2-根号3
所以y=x-(1/2+根号3)=z-(1/2-根号3)
z-x=-2倍根号3
带入(1/2)[(x-y)^2+(z-y)^2+(z-x)^2] 得出结果就好啦
应该是37/4吧

Answer 2

x-y=(1+根号3)/2
z-y=(1-根号3)/2
x-y-(z-y)=(1+根号3)/2-(1-根号3)/2
x-z=根号3
x²+y²+z²-xy-yz-xz
=(2x^2+2y^2+2z^2-2xy-2yz-2xz)/2
=[(x^2+y^2-2xy)+(y^2+z^2-2yz)+(z^2+x^2-2xz)]/2
=[(x-y)^2+(y-z)^2+(x-z)^2]/2
={[(1+根号3)/2]^2+[(1-根号3)/2]^2+(根号3)^2}/2
={[(4+2根号3)+(4-2根号3)]/4+3}/2
=(2+3)/2
=5/2