已知x^2-5x-2004=0,求代数式(x-2)^3-(x-1)^2+1除以x-2

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/27 21:47:07
已知x^2-5x-2004=0,求代数式(x-2)^3-(x-1)^2+1除以x-2

已知x^2-5x-2004=0,求代数式(x-2)^3-(x-1)^2+1除以x-2
已知x^2-5x-2004=0,求代数式(x-2)^3-(x-1)^2+1除以x-2

已知x^2-5x-2004=0,求代数式(x-2)^3-(x-1)^2+1除以x-2
[(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

已知x^2-5x-2004=0,求代数式[(x-2)^3-(x-1)^2+1]/(x-2)的值 =[(x^3-6x^2+12x-8)-(x^2-2x+1)+1]/(x-2) =(x^3-7x^