Xin phép sửa đề:
Ta có: \(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}=\frac{x+3}{1-x^2}\) \(\left(x\ne\pm1\right)\)
\(\Leftrightarrow\frac{\left(3x+1\right)\left(x+1\right)-\left(1-x\right)^2}{\left(1-x\right)^2\left(x+1\right)}=\frac{\left(x+3\right)\left(1-x\right)}{\left(1-x\right)^2\left(x+1\right)}\)
\(\Rightarrow3x^2+4x+1-1+2x-x^2=-x^2-2x+3\)
\(\Leftrightarrow3x^2+8x-3=0\)
\(\Leftrightarrow\left(3x^2+9x\right)-\left(x+3\right)=0\)
\(\Leftrightarrow3x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\3x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
Vậy tập nghiệm PT \(S=\left(-3;\frac{1}{3}\right)\)
