Question

已知x-y-3z=0,x+y-z=0,则（x^2 + y^2 - 3z^2 - xy + yz - zx）/(x^2 -

Answer 1

x-y-3z=0 1)

x+y-z=0 2)

1)+2),得

x=2z

y=-z

所以原式=(4z^2+z^2-3z^2+2x^2-z^2-2z^2)/(4z^2-3z^2+z^2-2z^2-z^2+2z^2)

=(4+1-3+2-1-2)/(4-3+1-2-1+2)

=1/1

=1