Question

若x^2-5x-1999=0,求[(x-2)^4+(x-1)^2-2]/[(x-1)(x-2)]

Answer 1

[(x-2)^4+(x-1)^2-1]/[(x-1)(x-2)]=[(x^2-4x+4)^2+(x-1)^2-1]/(x^2-3x+2)

=[(x^2-5x-1999+x+2003)^2+(x-1)^2-1]/(x^2-3x+2)

=[(x+2003)^2+(x-1)^2-1]/(x^2-3x+2)

=[x^2+4006x+2003^2+x^2-2x]/(x^2-3x+2)

=[2x^2+4004x+2003^2]/(x^2-3x+2)

=[2(x^2-5x-1999)+4014x+2003^2+2*1999]/(x^2-3x+2)

=[4014x+2003^2+2*1999]/(x^2-5x-1999+2x+2001)

=[4014x+4016007]/(2x+2001)

=2007[2x+2001]/(2x+2001)

=2007