Question
$$\frac{2}{x^2-1}+\frac{1}{1-x} = 0$$
Answer
$$ x = 1 $$
Explanation
$$ \begin{aligned} \frac{2}{x^2-1}+\frac{1}{1-x} &= 0&& \text{multiply ALL terms by } \color{blue}{ (x^2-1)\cdot(1-x) }. \\[1 em](x^2-1)\cdot(1-x)\cdot\frac{2}{x^2-1}+(x^2-1)\cdot(1-x)\cdot\frac{1}{1-x} &= (x^2-1)\cdot(1-x)\cdot0&& \text{cancel out the denominators} \\[1 em]-2x+2+x^2-1 &= 0&& \text{simplify left side} \\[1 em]x^2-2x+1 &= 0&& \\[1 em] \end{aligned} $$
$ x^{2}-2x+1 = 0 $ is a quadratic equation.
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