题目

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

提问时间：2021-09-10

答案

[(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