如题,x^2表示x的2次方,后一个代数式怎么化简?x^2-3x+1=0,求2x^5-5x^4-x^3+x^2-6x/x^2+1的值_数学解答

如题,x^2表示x的2次方,后一个代数式怎么化简?x^2-3x+1=0,求2x^5-5x^4-x^3+x^2-6x/x^2+1的值,

人气：360 ℃　时间：2019-10-08 08:34:06

解答

2x^5-5x^4-x^3+x^2-6x/(x^2+1)
=2x^5-5x^4-x^3+x^2-6x/3x
=2x^5-5x^4-x^3+x^2-2
=2x^3(x^2-3x+1)+x^4-3x^3+x^2-2
=2x^3*0+x^2(x^2-3x+1)-2
=x^2*0-2
=-2.