Điều kiện: x + 1 > 0 ⇒ x > -1
Mẫu chung là x.(x+1)
$\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x.(x+1)}$
⇔ $\frac{(x-1).(x+1)}{x.(x+1)}$ - $\frac{1x}{x.(x+1)}$ = $\frac{2x-1}{x.(x+1)}$
⇔ (x-1).(x+1) - x = 2x - 1
⇔ $x^{2}$ - 1 - x -2x +1 = 0
⇔ $x^{2}$ - 3x = 0
⇔ x.(x-3) = 0
⇔ \(\left[ \begin{array}{l}x=0 \\x=3\end{array} \right.\)