解
x=a是方程x²-2003x+1=0的根
则a²-2003a+1=0
∴a²+1=2003a
两边除以a
则a+1/a=2003
a²-2002a+2003/(a²+1)
=a²-2003a+a+2003/(a²+1)
=-1+a+2003/(2003a)
=a-1+1/a
=(a+1/a)-1
=2003-1
=2002