\(\dfrac{1}{x^2-1}+\dfrac{5}{x+1}-\dfrac{1}{x-1}=1\left(đk:x\ne\pm1\right)\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow1+5x-5-x-1=x^2-1\)
\(\Leftrightarrow x^2-5x+4=0\)
\(\Leftrightarrow x^2-x-4x+4=0\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình là \(S=\left\{4\right\}\)
