Question

Cho \(\frac{n}{n^2-n+1}\) = a. Tính \(\frac{n^2}{n^4+n^2+1}\) theo a

Answer 1

\(\frac{n}{n^2-n+1}=a\Leftrightarrow n=a\left(n^2-n+1\right)\)

\(\Leftrightarrow n^2=a^2\left(n^2-n+1\right)^2\)

\(\Leftrightarrow n^2=a^2\left(n^4+n^2+1-2n^3+2n^2-2n\right)\)

\(\Leftrightarrow n^2=a^2\left(n^4+n^2+1\right)-2a^2n\left(n^2-n+1\right)\)

\(\Leftrightarrow n^2=a^2\left(n^4+n^2+1\right)-2an^2\) ( vì \(a\left(n^2-n+1\right)=n\))

\(\Leftrightarrow n^2\left(2a+1\right)=a^2\left(n^4+n^2+1\right)\)

\(\Leftrightarrow\frac{n^2}{n^4+n^2+1}=\frac{a^2}{2a+1}\).