ĐKXĐ: x∉{0;1}
Ta có: \(\frac{x+1}{x-1}-\frac{3x+1}{x^2-x}=\frac{1}{x^2}\)
\(\Leftrightarrow\frac{x^2\left(x+1\right)}{x^2\left(x-1\right)}-\frac{x\left(3x+1\right)}{x^2\left(x-1\right)}=\frac{x-1}{x^2\left(x-1\right)}\)
Suy ra: \(x^2\left(x+1\right)-x\left(3x+1\right)=x-1\)
\(\Leftrightarrow x^3+x^2-3x^2-x-x+1=0\)
\(\Leftrightarrow x^3-2x^2-2x+1=0\)
\(\Leftrightarrow x^3+1-\left(2x^2+2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-2x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1-2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2-3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(x-\frac{3}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x-\frac{3}{2}=\frac{\sqrt{5}}{2}\\x-\frac{3}{2}=-\frac{\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=\frac{3+\sqrt{5}}{2}\left(tm\right)\\x=\frac{3-\sqrt{5}}{2}\left(tm\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\frac{3+\sqrt{5}}{2};\frac{3-\sqrt{5}}{2}\right\}\)
