\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=-\dfrac{x^2+3}{1-x^2}\)
ĐKXĐ: x ≠ 1; x ≠ - 1
\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=-\dfrac{x^2+3}{1-x^2}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^2+3}{x^2-1}\)
\(\Rightarrow\left(x+1\right)^2-\left(x-1\right)^2=x^2+3\)
\(\Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)-x^2-3=0\)
\(\Leftrightarrow4x-x^2-3=0\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
<=> (x - 1)(x - 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ko-thỏa-mản-đk\right)\\x=3\left(thỏa-mản-đk\right)\end{matrix}\right.\)
Vậy S = {3}
