$a.(5x-3)(x^2+4)(x-1)=0$
Vì $x^2≥0∀x$ nên $x^2+4>0∀x$
$⇒(5x-3)(x-1)=0$
$⇔\left[ \begin{array}{l}5x-3=0\\x-1=0\end{array} \right.⇔\left[ \begin{array}{l}x=\frac{3}{5}\\x=1\end{array} \right.$
Vậy $S=${$\frac{3}{5};1$}
$b.x(x^2-1)=0$
$⇔x(x+1)(x-1)=0$
$⇔\left[ \begin{array}{l}x=0\\x+1=0\\x-1=0\end{array} \right.⇔\left[ \begin{array}{l}x=0\\x=-1\\x=1\end{array} \right.$
Vậy $S=${$0;-1;1$}.