Câu a :
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\) \(\dfrac{1\left(2-x\right)}{\left(x+1\right)\left(2-x\right)}+\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(2-x\right)}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\) ( Loại )
Vậy \(S=\left\{\varnothing\right\}\)
Câu b :
ĐKXĐ : \(x\ne\pm2\)
\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
\(\Leftrightarrow\) \(\dfrac{\left(x-1\right)\left(2-x\right)}{\left(x+2\right)\left(2-x\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(2-x\right)}=\dfrac{5x-2}{\left(x+2\right)\left(2-x\right)}\)
\(\Leftrightarrow2x-x^2-2-x+x^2+2x=5x-2\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\) ( Loại )
Vậy \(S=\left\{\varnothing\right\}\)