\(\dfrac{x+2}{x-1}=\dfrac{x-1}{x-3}\) (1)
ĐKXĐ: \(x\ne1;x\ne3\)
(1) \(\Leftrightarrow\left(x+2\right)\left(x-3\right)=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-3x+2x-6=x^2-2x+1\)
\(\Leftrightarrow-3x+2x+2x=1+6\)
\(\Leftrightarrow x=7\) (nhận)
Vậy S = {7}