$\frac{x + 1}{x}$ + 1 = $\frac{3x - 1}{x + 1}$ + $\frac{1}{x(x+1)}$
ĐKXD: x$\neq$ 0
x$\neq$ -1
⇔ (x + 1)² + x² + x = 3x² - x + 1
⇔ x² + 2x + 1 + x² + x = 3x² - x + 1
⇔2x² - 3x² + 3x + x = 0
⇔ -x² + 4x = 0
⇔ x(-x + 4) =0
⇔\(\left[ \begin{array}{l}x = 0\\4 - x = 0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x = 0(koTMDK)\\x = 4(TMDK)\end{array} \right.\)
vậy S = { 4 }
học tốt
cho mik ctlhn ạ
