$(x+1+\frac{1}{x}) ^{2} = (x-1-\frac{1}{x}) ^{2}$ ($x \neq 0$)
$| x + 1 +\frac{1}{x}| = |x-1-\frac{1}{x} |$
\(\left[ \begin{array}{l}x + 1 +\frac{1}{x}=x-1-\frac{1}{x} \\x + 1 +\frac{1}{x}=-x+1+\frac{1}{x}\end{array} \right.\)
\(\left[ \begin{array}{l}x=-1\\x=0 (không thỏa)\end{array} \right.\)
Vậy $x=-1$