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