$\frac{1}{x - 2} + \frac{2}{x - 1} \cdot \frac{6}{x} = 0$

Answer

$x_{1} = - \frac{11}{2} - \frac{217}{2} x_{2} = - \frac{11}{2} + \frac{217}{2}$

Explanation

\frac{1}{x - 2} + \frac{2}{x - 1} \cdot \frac{6}{x} (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 + 12 x - 24 x^{2} + 11 x - 24 = 0 = (x - 2) (x - 1) x \cdot 0 = 0 = 0 multiply ALL terms by (x - 2) (x - 1) x . cancel out the denominators simplify left side

$x^{2} + 11 x - 24 = 0$ is a quadratic equation.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.

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x−21​+x−12​⋅x6​=0

x1​=−211​−2217​​​x2​=−211​+2217​​​

$\frac{1}{x - 2} + \frac{2}{x - 1} \cdot \frac{6}{x} = 0$

$x_{1} = - \frac{11}{2} - \frac{217}{2} x_{2} = - \frac{11}{2} + \frac{217}{2}$