ĐKXĐ:...
Đặt \(x+\frac{1}{x}=a\Rightarrow a^3=x^3+\frac{1}{x^3}+3x.\frac{1}{x}\left(x+\frac{1}{x}\right)\)
\(\Rightarrow a^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=x^3+\frac{1}{x^3}+3a\)
\(\Rightarrow x^3+\frac{1}{x^3}=a^3-3a\)
Thay vào pt ta được:
\(4\left(a^3-3a\right)=13a\)
\(\Leftrightarrow4a^3-25a=0\Leftrightarrow a\left(4a^2-25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\a=\frac{5}{2}\\a=-\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=0\\x+\frac{1}{x}=\frac{5}{2}\\x+\frac{1}{x}=-\frac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(vn\right)\\2x^2-5x+2=0\\2x^2+5x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm\frac{1}{2}\end{matrix}\right.\)
