已知x2-5x-2008=0,则代数式((x-2)3-(x-1)2+1)/(x-2)的值为A.2008 B.2010 C.2012 D.2014

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/30 06:17:44
已知x2-5x-2008=0,则代数式((x-2)3-(x-1)2+1)/(x-2)的值为A.2008 B.2010 C.2012 D.2014
x){}K+tM+t ,l tv|{ k)CM#mCM}Y-O<ٱQI H(8(#ebTO`3١_`gCsO<_;Y`vE-H" 0YgÓKh]qP+*@V` 7mTt!v@ITI&&)\`Y/:ut/.H̳ RO{=_` 0ѴX1aɏ

已知x2-5x-2008=0,则代数式((x-2)3-(x-1)2+1)/(x-2)的值为A.2008 B.2010 C.2012 D.2014
已知x2-5x-2008=0,则代数式((x-2)3-(x-1)2+1)/(x-2)的值为
A.2008 B.2010 C.2012 D.2014

已知x2-5x-2008=0,则代数式((x-2)3-(x-1)2+1)/(x-2)的值为A.2008 B.2010 C.2012 D.2014
答:C
x^2-5x-2008=0
x^2-5x=2008
所以:
[ (x-2)^3-(x-1)^2+1 ] /(x-2)
=(x-2)^2+(-x^2+2x-1+1) /(x-2)
=(x-2)^2+(2x-x^2) /(x-2)
=x^2-4x+4-x
=x^2-5x+4
=2008+4
=2012
选择C

2012选C
代数式化简=(x-1)(x-4)=x2-5x+4=2012