已知x2-5x-1991=0,则代数式(x−2)4+(x−1)2−1(x−1)(x−2)的值为(  ) A.1996 B.1997 C.1998 D.1999

问题描述:

已知x2-5x-1991=0,则代数式

(x−2)4+(x−1)2−1
(x−1)(x−2)
的值为(  )
A. 1996
B. 1997
C. 1998
D. 1999

(x−2)4+(x−1)2−1
(x−1)(x−2)
=
(x−2)4+x(x−2)
(x−1)(x−2)

=
(x−2)3+x
x−1

=
x3−6x2+12x−8+x
x−1

=
x2(x−1)−5x(x−1)+8(x−1)
x−1

=x2-5x+8;
∵x2-5x-1991=0,
∴x2-5x=1991,
∴原式=1991+8=1999.
故选D.