Mình thiếu điều kiện xác định ^_^
Cho mình bổ xung thêm
\(ĐKXĐ:x\ne\pm1\)
và mình sửa lại nữa là: \(\orbr{\begin{cases}x=-1\left(L\right)\\x=-3\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{-3\right\}\)
\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{x^2+3}{1-x^2}\) đkxđ \(x\ne\pm1\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{-x^2-3}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow x^2+2x+1-x^2-2x-1+x^2+3=0\)
\(\Leftrightarrow x^2+3=0\)
\(\Leftrightarrow x^2=-3\)
\(\Leftrightarrow x\in\varnothing\)
\(ĐKXĐ:x\ne\pm1\)
\(pt\Leftrightarrow\frac{\left(x+1\right)^2-\left(x-1\right)^2}{x^2-1}=\frac{-x^2-3}{x^2-1}\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=-x^2-3\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=-x^2-3\)
\(\Leftrightarrow x^2+4x+3=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-1\end{cases}}\)
Ta có: \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{x^2+3}{1-x^2}\)
\(\Leftrightarrow\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right).\left(x-1\right)}=-\frac{x^2+3}{x^2-1}\)
\(\Leftrightarrow\frac{\left[\left(x+1\right)-\left(x-1\right)\right].\left[\left(x+1\right)+\left(x-1\right)\right]}{x^2-1}=\frac{-\left(x^2+3\right)}{x^2-1}\)
\(\Rightarrow\left(x+1-x+1\right).\left(x+1+x-1\right)=-\left(x^2+3\right)\)
\(\Leftrightarrow2.2x=-x^2-3\)
\(\Leftrightarrow x^2+4x+3=0\)
\(\Leftrightarrow\left(x^2+x\right)+\left(3x+3\right)=0\)
\(\Leftrightarrow x.\left(x+1\right)+3.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
Vậy \(S=\left\{-1;-3\right\}\)
