Question

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

Answer 1

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