若x^2-x-1=0,则2x^3-4x^2+2005=?

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若x^2-x-1=0,则2x^3-4x^2+2005=?
若x^2-x-1=0,则2x^3-4x^2+2005=?

若x^2-x-1=0,则2x^3-4x^2+2005=?
2x^3-4x^2+2005
=2x(x^2-x-1)-2x^2+2x+2005
=-2x^2+2x+2+2003
=-2(x^2-x-1)+2003
=2003

解 因为x2=1+x 所以原式=-2x2+2x+2005=-2x-2+2x+2005=2003 降次即可。

x^2-x-1=0
x^2-x=1
2x^3-4x^2+2005
=2x^3-2x^2-2x^2+2005
=2x(x^2-x)-2x^2+2005
=2x-2x^2+2005
=-2(x^2-x)+2005
=-2*1+2005
=-2+2005
=2003