$$\frac{1}{x-2}+\frac{2}{x-1}\cdot\frac{6}{x} = 0$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 11 }{ 2 }-\dfrac{\sqrt{ 217 }}{ 2 } & x_2 = -\dfrac{ 11 }{ 2 }+\dfrac{\sqrt{ 217 }}{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{x-2}+\frac{2}{x-1}\cdot\frac{6}{x} &= 0&& \text{multiply ALL terms by } \color{blue}{ (x-2)(x-1)x }. \\[1 em](x-2)(x-1)x\cdot\frac{1}{x-2}+(x-2)(x-1)x\frac{2}{x-1}\cdot\frac{6}{x} &= (x-2)(x-1)x\cdot0&& \text{cancel out the denominators} \\[1 em]x^2-x+12x-24 &= 0&& \text{simplify left side} \\[1 em]x^2+11x-24 &= 0&& \\[1 em] \end{aligned} $$
$ x^{2}+11x-24 = 0 $ is a quadratic equation.
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