Question

已知x^-5x-1997=0,求代数式[（x-2)（x-2)(x-2)-(x-1)(x-1)+1]/(x-2)的值...

Answer 1

因为[(x-2)^3-(x-1)^2+1]/(x-2)

=[(x-2)^3/(x-2)]-[(x-1)^2-1]/(x-2)

=(x-2)^2-[(x-1+1)(x-1-1)/(x-2)]

=x^2-4x+4-x

=x^2-5x+4,

又因为x^2-5x-1997=0,

所以x^2-5x=1997,

所以原式=[(x-2)^3-(x-1)^2+1]/(x-2)

=x^2-5x+4

=1997+4=2001.