Bài giải:

$x+\frac{1}{x}={{x}^{2}}+\frac{1}{{{x}^{2}}}$

ĐKXĐ: x $\ne $0

$\Leftrightarrow \frac{{{x}^{2}}+1}{x}=\frac{{{x}^{4}}+1}{{{x}^{2}}}$

$\Leftrightarrow \frac{{{x}^{3}}+x}{{{x}^{2}}}=\frac{{{x}^{4}}+1}{{{x}^{2}}}$

$\Leftrightarrow $x³ + x = x⁴ + 1

$\Leftrightarrow $x⁴ – x³ – x + 1 = 0

$\Leftrightarrow $x³(x – 1) – (x – 1) = 0

$\Leftrightarrow $(x³ – 1)(x – 1) = 0

$\Leftrightarrow $(x – 1)² (x² + x + 1) = 0

Mà: x² + x + 1 $\ne $0 nên (x – 1)² = 0

$\Leftrightarrow $x = 1    (thỏa mãn)

Vậy x = 1