a.(5x – 3)(x^2 + 4)(x – 1) = 0 b. x(x^2 – 1) = 0 Phương trình tích

$a.(5x-3)(x^2+4)(x-1)=0$

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 $S=${$0;-1;1$}.