\(u+\dfrac{1}{u}=\dfrac{u^2+1}{u}=\dfrac{\left(\sqrt{2}+1\right)^2+1}{\sqrt{2}+1}\)
\(=\dfrac{4+2\sqrt{2}}{\sqrt{2}+1}=\dfrac{2\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\sqrt{2}\)
\(P=u^8+\dfrac{1}{u^8}\)
\(=\left(u^4+\dfrac{1}{u^4}\right)^2-2\)
\(=\left[\left(u^2+\dfrac{1}{u^2}\right)^2-2\right]^2-2\)
\(=\left\{\left[\left(u+\dfrac{1}{u}\right)^2-2\right]^2-2\right\}^2-2\)
\(=\left\{\left[\left(2\sqrt{2}\right)^2-2\right]^2-2\right\}^2-2\)
\(=1154\) (tại vì t lười tính nên khúc này bn tính tay hoặc bấm máy tính đều đc ^^)
u = \(\sqrt{2}+1\) => \(u^2=3+2\sqrt{2}\) => \(u^4=17+12\sqrt{2}\)=> \(u^8=577+408\sqrt{2}\) => \(u^{16}=665857+665857=1331714\)
P = \(u^8+\dfrac{1}{u^8}=\dfrac{u^{16}+1}{u^8}=\dfrac{1331714}{577+408\sqrt{2}}=1154\)
