数学题：已知x-y-3z=0,x+y-z=0,则

数学题：已知x-y-3z=0,x+y-z=0,则

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

解：已知x-y-3z=0 x+y-z=0 由上式+下式，得2x-4z=0 x=2z 由下式-上式，得2y+2z=0 y=-z 则：(x^2+y^2-3z^2-xy+yz-zx)/(x^2-3y^2+z^2+xy+yz+zx) =(4z^2+z^2-3z^2+2z^2-z^2-2z^2)/(4z^2-3z^2+z^2-2z^2-z^2+2z^2) =z^2/z^2=1

x-y-3z=0,x+y-z=0, x=2z,y=-z 代入就是了 再看看别人怎么说的。

前两个式子相加，求出2x=4z x=2z 带入求出y=-z 然后带入后面那个式子 原式=(4z^2+z^2-3z^2+2z^2-z^2-2z^2)/(4z^2-3z^2+z^2-2z^2-z^2+2z^2) =z^2/z^2=1

x-y-3z=0,x+y-z=0, x=2z,y=-z 代入就是了