*if x^2-5x+1=0...數學題一問.20分

x^2-5x+1=0
(x^2-5x+1)/x^2 = 0
1 - 5/x + 1/x^2 = 0
- 5/x + 1/x^2 = - 1
5/x - 1/x^2 = 1 ......(*)
;
x^2/(x^4+5x^2+1)
= (x^2) / [ (x^2 - 5x + 1)(x^2 + 5x + 29) + 140x - 28 ]
= (x^2) / (140x - 28)
= 1 / [(140x - 28)/x^2]
= 1 / [28(5x - 1) / x^2]
= 1 / [28 (5/x - 1/x^2)] ,
by (*)
= 1 / [28(1)]
= 1/28


2010-07-11 12:22:40 補充：
Method 2 :

x^2-5x+1=0

x^2 = 5x - 1 ......(*)

;

x^2/(x^4+5x^2+1)

= (x^2) / [ (x^2 - 5x + 1)(x^2 + 5x + 29) + 140x - 28 ]

= (x^2) / (140x - 28)

= (x^2) / [28(5x - 1)] , by (*)

= (x^2) / (28x^2)

= 1/28

雖然我覺得係你打錯字
但, 人手計唔到

x=0.2087

x^2/(x^4+5x^2+1)=0.035712