已知x/x²+x+1=a,试求分式x²/x四次方+x²+1
问题描述：
已知x/x²+x+1=a,试求分式x²/x四次方+x²+1
答
x/x²+x+1=a
1/x+x=a-1
x²/x四次方+x²+1=1/x²+x²+1=(1/x+x)²-2+1
代入
（a-1)²-1=a²-2a
答
x/(x²+x+1)=a
(x²+x+1)/x=1/a=x+1/x+1
x+1/x=1/a-1
x²/x四次方+x²+1
=1/(x^2+1/x^2+1)
=1/[(x+1/x)^2-1]
=1/[(1/a-1)^2-1]
=a^2/(1-2a)