Bạn tham khảo nha:
$x$+$\frac{1}{x}$ =$x^{2}$+$\frac{1}{x^2}$ (ĐKXĐ x khác 0)
⇔ $\frac{x^2+1}{x}$= $\frac{x^4+1}{x^2}$
⇔$x^{2}$.($x^{2}$+$1$)= ($x^{4}$+$1$).$x$
⇔$x^{4}$+$x^{2}$= $x^{5}$+$x$
⇔$x^{4}$+$x^{2}$- $x^{5}$-$x$ =0
⇔($x^{4}$-$x^{5}$)-($x$-$x^{2}$)=0
⇔$x^{4}$.$(1-x)$-$x$.$(1-x)$ =0
⇔($x^{4}$-$x$).$(1-x)$=0
⇔$(x-1)$.($x^{2}$+x+1).$(1-x)$=0
⇔\(\left[ \begin{array}{l}x=0(ko thõa mãn)\\1-x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0(ko thõa mãn)\\x=1\end{array} \right.\)
