Question

x的平方-5x-2004=0，求（x-2）的3次方-（x-1)的2次方+1除以x-2的值

Answer 1

解答：
[(x-2)^3-(x-1)^2+1]/(x-2)=[x^3-6x^2+12x-8-x^2+2x-1+1]/(x-2)
=[x^3-7x^2+14x-8]/(x-2)=[(x-2)(x^2+2x+4)-7x(x-2)]/(x-2)
=x^2-5x+4=2004+4=2008

Answer 2

x^2-5x-2004=0
( x-2)^2-x-2008=0
( x-2)^2=x+2008
（x-2）^3-（x-1)^2+1
=(x+2008)(x-2)-x^2+2x-1+1
=x^2+2006x-2008*2--x^2+2x
=2008x-2008*2
=2008(x-2)
所以[（x-2）^3-（x-1)^2+1]/(x-2)=2008