设方程2002^2* x^2-2003*2001x-1=0的较大根为r,方程2001x^2-2002x+1=0的最小根为s,则r-s的值为____

有个公式x^2-y^2=(x+y)(x-y)
1可以看作1^2
方程一因式分解(2002^2x+1)(x-1)=0
r=1
方程二因式分解(2001x-1)(x-1)=0
s=1/2001
r-s=2000/2001
给我点分吧

2002^2* x^2-2003*2001x-1=0
(2002x)^2-2003×2001x-1=0
(2002x)^2-(2002+1)×(2002-1)x-1=0
2002^2x^2-(2002^2-1)x-1=0
2002^2x^2-2002^2x+x-1=0
2002^2(x-1)x+(x-1)=0
(2002^2x+1)(x-1)=0
x1=-1/2002^2
x2=1

2001x^2-2002x+1=0
(2001x-1)(x-1)=0
x1=1/2001
x2=1

r-s
=1-1/2001
=2000/2001