Question

设x/x^2+x+1=1/5,求x^2/x^4+x^2+1的值 大虾们帮下忙,3q~

Answer 1

设x/(x^2+x+1)=1/5,求x^2/(x^4+x^2+1)的值 大虾们帮下忙

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

=>两边倒数

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

=>分式分解

x+1+1/x=5

=>两边减一

x+1/x=4

=>两边平方

x^2+2+1/x^2=16

=>两边减二

x^2+1/x^2=14

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

=1/[(x^4+x^2+1)/x^2]

=1/[x^2+1+1/x^2]

=1/[(x^2+1/x^2)+1]

=1/[14+1]

=1/15