Question

已知x²－5x＋1＝0求x四次方－x²＋1分之x²

Answer 1

x²-5x+1=0
显然x≠0，两边同时x得到：x-5+(1/x)=0
==> x+(1/x)=5
两边平方得到：[x+(1/x)]²=5²=25
==> x²+2x*(1/x)+(1/x)²=25
==> x²+(1/x)²=23
所以，x²/(x^4-x²+1)=1/[x²-1+(1/x)²]=1/{[x²+(1/x²)]-1}=1/(23-1)=1/22

Answer 2

等式两边同除以x，得：
x- 5 +1/x=0
x+ 1/x=5
x²/(x⁴-x²+1)
=1/(x² -1 +1/x²) (分子分母同除以x²)
=1/(x²+2+ 1/x² -3)
=1/[(x+ 1/x)²-3]
=1/(5²-3)
=1/(25-3)
=1/22

Answer 3

x^2-5x+1=0
x+1/x =5
(x+1/x)^2=25
x^2+1/x^2 =23

x^2/(x^4-x^2+1)
=1/(x^2-1 + 1/x^2)
=1/(23-1)
=1/22

Answer 4

解原式=x^2/x^4-x^2+1
=1/x^2+1/x^2-1
因为x^2-5x+1=0
所以x+1/x=5
(x+1/x)^2=25
x^2+1/x^2+2=25
x2+1/x^2=23 (1)
将（1）代人原式得
=1/(23-1)
=1/22
所以所求代数式的值是1/22