ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x}{x-2}+1=\dfrac{2x^2+3}{x^2-4}\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+3}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+2x+x^2-4=2x^2+3\)
\(\Leftrightarrow2x-4=3\)
\(\Leftrightarrow2x=7\)
\(\Leftrightarrow x=\dfrac{7}{2}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{7}{2}\right\}\)