已知x2-5x-2000=0,则代数式(x−2)3−(x−1)2+1x−2的值为_.

问题描述:

已知x2-5x-2000=0,则代数式

(x−2)3(x−1)2+1
x−2
的值为______.

∵x2-5x-2000=0,
∴x2-5x=2000.
又∵(x-2)3-(x-1)2+1
=(x-2)3-[(x-1)2-1]
=(x-2)3-[(x-1+1)(x-1-1)]
=(x-2)3-x(x-2)
=(x-2)[(x-2)2-x]
=(x-2)(x2-5x+4),

(x−2)3(x−1)2+1
x−2
=
(x−2)(x2−5x+4)
x−2
=x2-5x+4=2000+4=2004.
故答案为:2004.