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-x

Answer 1

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)]/...