$$1 = \frac{1}{x^2+2x}+\frac{x-1}{x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -1 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 1 &= \frac{1}{x^2+2x}+\frac{x-1}{x}&& \text{multiply ALL terms by } \color{blue}{ (x^2+2x)x }. \\[1 em](x^2+2x)x\cdot1 &= (x^2+2x)x\cdot\frac{1}{x^2+2x}+(x^2+2x)x\frac{x-1}{x}&& \text{cancel out the denominators} \\[1 em]x^3+2x^2 &= x+x^3+x^2-2x&& \text{simplify right side} \\[1 em]x^3+2x^2 &= x^3+x^2-x&& \text{move all terms to the left hand side } \\[1 em]x^3+2x^2-x^3-x^2+x &= 0&& \text{simplify left side} \\[1 em]x^3+2x^2-x^3-x^2+x &= 0&& \\[1 em]x^2+x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{2}+x = 0 } $, first we need to factor our $ x $.
$$ x^{2}+x = x \left( x+1 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ x+1 = 0$.
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