已知x^2-5x-2010=0则分式[(x-2)^4+(x-1)^2-1]/(x-1)(x-2)的值为_数学解答

已知x^2-5x-2010=0则分式[(x-2)^4+(x-1)^2-1]/(x-1)(x-2)的值为

人气：106 ℃　时间：2020-04-21 05:20:47

解答

令a=x-2
x-1=a+1
原式=[a^4+(a+1)²-1]/a(a+1)
=(a^4+a²+2a)/a(a+1)
=a(a³+1+a+1)/a(a+1)
=[(a+1)(a²-a+1)+(a+1)]/(a+1)
=(a+1)(a²-a+1+1)/(a+1)
=a²-a+2
=(x-2)²-(x-2)+2
=x²-5x+8
=2010+8
=2018