c)
ĐKXĐ: \(x\notin\left\{1;-3\right\}\)
Ta có: \(\dfrac{2x-1}{x-1}-\dfrac{2x+5}{x+3}=\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{2x^2+6x-x-3}{\left(x-1\right)\left(x+3\right)}-\dfrac{2x^2-2x+5x-5}{\left(x+3\right)\left(x-1\right)}=\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
Suy ra: \(2x^2+5x-3-2x^2-3x+5=4\)
\(\Leftrightarrow2x+2=4\)
\(\Leftrightarrow2x=2\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
