Question

已知x/(x^2+x+1)=a,求分式x^2/(x^4+x^2+1)的值

用取倒的方法 急！急！急！
谢谢

Answer 1

x/(x^2+x+1)=a,
1/[x+1+(1/x)]=a,
x+(1/x)+1=1/a,
x+(1/x)=(1-a)/a,

x^2/(x^4+x^2+1)
=1/[x^2+1+(1/x^2)]
=1/{[x+(1/x)]^2-2*x*(1/x)+1}
=1/{[(1-a)/a]^2-1}
=1/[(1-a)^2/a^2-a^2/a^2]
=1/[(1-2a)/a^2]
=a^2/(1-2a).