Đáp án:
$\begin{array}{l}
a)\left( {2{x^2} + 1} \right)\left( {{x^2} - 1} \right)\left( {{x^2} - 9} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
2{x^2} + 1 = 0\left( {vô\,nghiệm} \right)\\
{x^2} - 1 = 0\\
{x^2} - 9 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} = 1\\
{x^2} = 9
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 1\\
x = - 1\\
x = 3\\
x = - 3
\end{array} \right.\\
b)\left( {{x^2} + 2x + 3} \right)\left( {{x^2} - 25} \right)\left( {x + 19} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + 2x + 3 = 0\\
{x^2} = 25\\
x = - 19
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{\left( {x + 1} \right)^2} + 2 = 0\left( {vn} \right)\\
x = - 5\\
x = 5\\
x = - 19
\end{array} \right.\\
Vậy\,x = - 5;x = 5;x = - 19
\end{array}$
