Question

已知x-y=1,y-z=3,求代数式x^2+y^2+z^2-xy-yz-zx

Answer 1

x-y=1,

y-z=3

两式相加得

x-z=4

x^2+y^2+z^2-xy-yz-zx

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

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

=1/2[(x-y)^2+(y-z)^2+(x-z)^2]

=1/2[1^2+3^2+4^2]

=1/2*26

=13