Question

已知x2+3x+1=0,求x2/x4+3x2+1的值

Answer 1

1/10
x4+3x2+1=x4-x3+(x3+3x2+x)-x+1=x4-x3+x(x2+3x+1)-x+1=x4-x3-x+1
=x4-(x3+3x2+x)+3x2+1=x4-x(x2+3x+1)+3x2+1=x4+3x2+1
=(x4+3x3+x2)-3x3+2x2+1=x2(x2+3x+1)-3x3+2x2+1=-3x3+2x2+1
=(-3x3-9x2-3x)+11x2+3x+1=-3x(x2+3x+1)+11x2+3x+1=10x2+x2+3x+1=10x2
所以原式=x2/10x2=1/10

Answer 2

x2+3x+1=0是指：x^2+3x+1=0 吧
∵ x^2+3x+1=0
∴ x^2=-3x-1
∴ x^2/x^4+3x^2+1
=1/x^2 + 3 x^2 + 1
=1/(-3x-1)+3(-3x-1)+1
=2-9x-1/(3x+1)
解方程x^2+3x+1=0，得：x=(±√5 - 3)/2
代入上式，得：
原式=2-9(±√5 - 3)/2-1/(±√5 - 3)/2

后面自己整理