$\frac{2x+1}{x-1}$ = $\frac{5(x-1)^2}{x+1}$ Đkxđ : x $\neq$ ± 1
=> (2x+1)(x+1)=$5(x-1)^{2}$
<=> 2$x^{2}$ +2x+x+1=5($x^{2}$ -2x+1)
<=> $2x^{2}$ -$5x^{2}$ +3x+10x+1-5=0
<=> $-3x^{2}$ +13x-4=0
<=> $-3x^{2}$ +12+x-4=0
<=> -3x(x-4)+(x+4 ) =0
<=> (x-4)(-3x+1)=0
<=> \(\left[ \begin{array}{l}x-4=0\\-3x+1=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=4\\x=\frac{1}{3}\end{array} \right.\)
Vậy S = {4; $\frac{1}{3}$ }
