(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．