Question

x+y+z=0,x^2+y^2+3z^2=1

已知x+y-z=0,x-y+3z=0,求(x^2+2y^2+3z^2)/(2x^2+y^2+3z^2)的值

Answer 1

x+y-z=0 ① →z=x+y③
x-y+3z=0 ②
①*3+② →2x+y=0 →y=-2x④
把④代入③ z=-x
x^2+2y^2+3z^2)/(2x^2+y^2+3z^2)
=(x*x+8x*x+3x*x)/(2x*x+4x*x+3x*x)
=12/9
=4/3