设x=2003
1+(2003-1)^2+(1-1/2003)^2
=1+(x-1)^2+(1-1/x)^2
=[(-x)^4+2(-x)^3+3(-x)^2+2(-x)+1]/x^2
=(1-x+x^2)^2/x^2
√[1+(2003-1)^2+(1-1/2003)^2]-1/2003
=(1-2003+2003^2)/2003-1/2003=2002