已知x^2-5x-2006=0,求代数式<(x-2)^3-(x-1)^2+1>⼀(x-2)的值

2024-11-23 23:38:10
推荐回答(2个)
回答1:

x^2-5x-2006=0
x^2-5x=2006
<(x-2)^3-(x-1)^2+1>/(x-2)
=[(x-2)^3+1-(x-1)^2]/(x-2)
=[(x-2)^3+(1-x+1)(1+x-1)]/(x-2)
=[(x-2^3)-x(x-2)]/(x-2)
=(x-2)^2-x
=x^2-5x+4
=2006+4
=2010

回答2:

原式=(x-2)^2+[1-(x-1)^2]/(x-2)=(x-2)^2+x(1-x+1)/(x-2)=x^2-4x+4-x=(x^2-5x-2005)+(4+2005)=0+4+2005=2009