初二数学分式题：已知x-y-z=0,y-z=0,且xyz≠0.

y-z=0,y=z
x-y-z=0,x=y+z=2z

(1998x^2+1999y^2-2000z^2)／(1998x^2-1999y^2+2000z^2)
=(1998*4z^2+1999z^2-2000z^2)/(1998*4z^2-1999z^2+2000z^2
=(1998*4-1)/(1998*4+1)
=7991/7993

y-z=0,y=z
x-y-z=0,x=y+z=2z

(1998x^2+1999y^2-2000z^2)／(1998x^2-1999y^2+2000z^2)
=(1998*4z^2+1999z^2-2000z^2)/(1998*4z^2-1999z^2+2000z^2
=(1998*4-1)/(1998*4+1)
=7991/7993
7991/7993是(1998x^2+1999y^2-2000z^2)／(1998x^2-1999y^2+2000z^2) 的值。

x-y-z=0
x=y+z
y-z=0
y=z
所以x=y+y=2y
y=z=x/2
y^2=z^2=x^2/4

(1998x^2+1999y^2-2000z^2)/(1998x^2-1999y^2+2000z^2)
=(1998x^2+1999*x^2/4-2000*x^2/4)/(1998x^2-1999*x^2/4+2000*x^2/4)
=(1998+1999/4-2000/4)/(1998-1999/4+2000/4)
=(1998-1/4)/(1998+1/4)
=(4*1998-1)/(4*1998+1)
=7991/7993

1/x+1/y=1/2
(x+y)/xy=1/2
xy=2(x+y)

所以(3x-5xy+3y)/(-x+3xy-y)
=[3(x+y)-5*2(x+y)]/[-(x+y)+3*2(x+y)]
=-7(x+y)/[