已知x2-5x-1991=0,则代数式(x−2)4+(x−1)2−1(x−1)(x−2)的值为(  )
1个回答

解题思路:首先要化简分式到最简,再把已知条件变形,代入即可.

(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.

点评:

本题考点: 分式的化简求值.

考点点评: 解答此题的关键是把分式化到最简,这个过程难度较大.