ĐKXĐ : \(x\ne0\)
Ta có : \(x-\frac{1}{x}=4\)
=> \(x^2-1=4x\)
=> \(x^2-4x-1=0\)
=> \(x^2-2.2x+4-5=0\)
=> \(\left(x-2\right)^2=5\)
=> \(x=2\pm\sqrt{5}\left(TM\right)\)
Ta có : \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}F=\frac{\left(2-\sqrt{5}\right)^2}{\left(2-\sqrt{5}\right)^4+1}\approx0,06\\F=\frac{\left(2+\sqrt{5}\right)^2}{\left(2+\sqrt{5}\right)^4+1}=\frac{1}{18}\end{matrix}\right.\\\left\{{}\begin{matrix}G=\left(2-\sqrt{5}\right)^4+\frac{1}{\left(2-\sqrt{5}\right)^4}=322\\G=\left(2+\sqrt{5}\right)^4+\frac{1}{\left(2+\sqrt{5}\right)^4}=322\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
\(x-\frac{1}{x}=4\Rightarrow\left(x-\frac{1}{x}\right)^2=16\Rightarrow x^2+\frac{1}{x^2}=18\)
\(\frac{1}{F}=\frac{x^4+1}{x^2}=x^2+\frac{1}{x^2}=18\Rightarrow F=\frac{1}{18}\)
\(\left(x^2+\frac{1}{x^2}\right)^2=324\Rightarrow x^4+\frac{1}{x^4}=322\)
\(\Rightarrow G=322\)
