a, $\dfrac{5}{x+2}=3$ (ĐKXĐ: $x \neq -2$)
$↔\dfrac{5}{x+2}=\dfrac{3.(x+2)}{x+2}$
$→5=3x+6$
$↔x=\dfrac{-1}{3}$ (TMĐK)
Vậy:`S={\frac{-1}{3}}`
b, $\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}$ (ĐKXĐ: $x \neq 0$)
$↔\dfrac{3x}{8x^2}-\dfrac{4x}{8x^2}=\dfrac{8}{8x^2}$
$→3x-4x=8$
$↔x=-8$ (TMĐK)
Vậy: `S={-8}`