$$\frac{x}{x-3}\frac{x-1}{2x} = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 1 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{x}{x-3}\frac{x-1}{2x} &= 0&& \text{multiply ALL terms by } \color{blue}{ (x-3)\cdot2x }. \\[1 em](x-3)\cdot2x \cdot \frac{x}{x-3}\frac{x-1}{2x} &= (x-3)\cdot2x\cdot0&& \text{cancel out the denominators} \\[1 em]x^4-x^3 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{4}-x^{3} = 0 } $, first we need to factor our $ x^3 $.
$$ x^{4}-x^{3} = x^3 \left( x-1 \right) $$$ x = 0 $ is a root of multiplicity $ 3 $.
The second root can be found by solving equation $ x-1 = 0$.
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