已知x平方-5x-2007=0,求(x-2)的立方-(x-1)的平方+1 x-2

问题描述:

已知x平方-5x-2007=0,求(x-2)的立方-(x-1)的平方+1 x-2
x-2前的空格是分号

因为X^2-5X-2007=0,即X^2-5X=2007所以 [(X-2)^3-(X-1)^2+1]/(X-2)=(x-2)^2-(x^2-2x+1)/(x-2)+1/(x-2) =(x-2)^2-(x^2-2x)/(x-2)-1/(x-2)+1/(x-2) =x^2-4x+4-x- 1/(x-2)+1/(x-2) =x^2-5x+4 =2007+4 =2011