$$\frac{1}{x-1}+\frac{1}{x+1} = \frac{1}{x}$$
Answer
$$ \begin{matrix}x_1 = i & x_2 = -i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{x-1}+\frac{1}{x+1} &= \frac{1}{x}&& \text{multiply ALL terms by } \color{blue}{ (x-1)(x+1)x }. \\[1 em](x-1)(x+1)x\cdot\frac{1}{x-1}+(x-1)(x+1)x\cdot\frac{1}{x+1} &= (x-1)(x+1)x\cdot\frac{1}{x}&& \text{cancel out the denominators} \\[1 em]x^2+x+x^2-x &= x^2-1&& \text{simplify left side} \\[1 em]2x^2 &= x^2-1&& \text{move all terms to the left hand side } \\[1 em]2x^2-x^2+1 &= 0&& \text{simplify left side} \\[1 em]x^2+1 &= 0&& \\[1 em] \end{aligned} $$
$ x^{2}+1 = 0 $ is a quadratic equation.
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