已知x/(x^2-4x-1)=-1/3,求x^4+2x+1/x^5的值

已知x/(x^2-4x-1)=-1/3,求x^4+2x+1/x^5的值
已知x/(x^2-4x-1)=-1/3,求x^4+2x+1/x^5的值

已知x/(x^2-4x-1)=-1/3,求x^4+2x+1/x^5的值
x/(x^2-4x-1)=-1/3
x^2-4x-1=-3x
x^2-x-1=0
x^2=x+1
x^4=x^2+2x+1=x+1+2x+1=3x+2
x^5=x^4*x=(3x+2)x=3x^2+2x=3(x+1)+2x=5x+3
(x^4+2x+1)/x^5
=(3x+2+2x+1)/(5x+3)
=(5x+3)/(5x+3)
=1

由已知变形可得x^2=x+1
所求式分子x^4+2x+1=（x+1）^2+2x+1=x^2+4x+2=（x+1）+4x+2=5x+3
分母x^5=（x+1）^2*x=（x^2+2x+1）*x=（x+1+2x+1）*x=3x^2+2x=3（x+1）+2x=5x+3
所以原式=1