a+1/a=3,求a^2/(a^4+a^2+1)

来源:百度知道 编辑:UC知道 时间:2024/06/23 14:57:42
已知a+1/a=3,求a^2/(a^4+a^2+1)

a+1/a=3
两边平方
a^2+1/a^2=7
a^2/(a^4+a^2+1)
=1/(1+a^2+1/a^2)
=1/8

0.125

a+1/a=3, 两边平方: a^2+1/a^2+2*a*1/a=a^2+1/a^2+2=3^2=9

a^2+1/a^2=8-2=7

a^2/(a^4+a^2+1) = 1/ (a^2+1/a^2+1)=1/(7+1)=1/8

因a+1/a=3
则两边平方得
(a+1/a)^2=3^2
即a^2+(1/a)^2+2*a*(1/a)=3^2
即a^2+1/(a^2)+2=9
即a^2+1/(a^2)=7

a^2/(a^4+a^2+1)
=(a^2/a^2)/[(a^4+a^2+1) a^2]
=1/(a^2+1+1/a^2)
=1/[(a^2+1/a^2)+1]
=1/[7+1]
=1/8

a+1/a=3
两边平方
a^2+1/a^2=7
a^2/(a^4+a^2+1)
=1/(1+a^2+1/a^2)
=1/8

这题先求出a^2/(a^4+a^2+1) 的倒数的方法更妙吧?
(a^4+a^2+1)/a^2
=a^2+1+1/a^2
=(a+1/a)^2-1
=8
所以a^2/(a^4+a^2+1)=1/8