Question

已知x^2-3x+1=0，求x^4+x^2+1分之x^2的ŀ

Answer 1

x^2/(x^4+x^2+1)
=(3x-1)/((3x-1)^2+(3x-1)+1)
=(3x-1)/(9x^2-3x+1)
=(3x-1)/(9(3x-1)-3x+1)
=(3x-1)/(24x-8)
=1/8

Answer 2

x²-3x+1=0，易知x≠0，两边同时除以x：
x-3+1/x=0
x+1/x=3
所求式子的倒数为：
(x⁴+x²+1)/x²
=x²+1+1/x²
=(x²+2+1/x²)-1
=(x+1/x)²-1
=3²-1
=9-1
=8
因此，所求式子的值为：1/8