Question

已知x^2－3x+1=0,求x^2+1/x^2与x^2/x^4+3x^2+1的值。

两个计算过程分开写，谢谢。。~~~~

Answer 1

x^2－3x+1=0
x^2+1=3x
因为x≠0
所以x+1/x=3

x^2+1/x^2
=(x+1/x)^2-2
=3^2-2
=7

x^2/x^4+3x^2+1
上下除以x^2
=1/(x^2+3+1/x^2)
=1/(7+3)
=1/10

Answer 2

x²-3x+1=0
显然x≠0，所以两边除以x
x-3+1/x=0
x+1/x=3
两边平方得
x²+2+1/x²=9
x²+1/x²=7

x²/(x^4+3x^2+1)
=1/(x²+3+1/x²)
=1/[(x²+1/x²)+3]
=1/(7+3)
=1/10

Answer 3

x^2－3x+1=0
x-3+1/x=0
x+1/x=3

x^2+1/x^2=(x+1/x)^2-2=3^2-2=7

x^2/x^4+3x^2+1
=1/(x^2+3+1/x^2)
=1/(7+3)
=1/10