∵x-5x-1999=0 ∴x^2-5x=1999 [(x-2)-（x-1)+1]/(x-2) =[(x-2)+1-（x-1)]/(x-2) ={(x-2+1)[(x-2)^2-(x-2)+1]-(x-1)^2}/(x-2) =[(x-1)(x^2-5x+7)-(x-1)^2]/(x-2) =(x-1)(x-2)(x-4)/(x-2) =(x-1)(x-4) =x^2-5x+4 =1999+4 =2003