若x^2-3x+1=0,则(2x^5-5x^4+2x^3-8x^2)/(x^2+1)的值是多少?
问题描述:
若x^2-3x+1=0,则(2x^5-5x^4+2x^3-8x^2)/(x^2+1)的值是多少?
答
x^2-3x+1=0
则x^2=3x-1,1=3x-x^2,x^+1=3x(代入下面的式子)
(2x^5-5x^4+2x^3-8x^2)/(x^2+1)
=[ 2x^3(x^-3x+1)+x^4-8x^2 ]/(x^2+1)
=[2x^2*0+(3x-1)^2-8x^2 ]/(3x-1+1)
=[9x^2-6x+1-8x^ ]/3x
= [ x^2-6x+1]/3x
=[3x-6x] /3x
=-3x/3x=-1
答
∵x^2-3x+1=0
∴x^2+1=3X
∴(2x^5-5x^4+2x^3-8x^2)/(x^2+1)
=[2X^3(X^2-3x+1)+x^2(x^2-3x+1)+3x(x^2-3x+1)-3x]/3x
=-3x/3x
=-1