Đáp án:

$\begin{array}{l}
a)x\left( {x - 1} \right)\left( {x + 1} \right)\left( {x + 2} \right) = 24\\
 \Leftrightarrow x\left( {x + 1} \right)\left( {x - 1} \right)\left( {x + 2} \right) = 24\\
 \Leftrightarrow \left( {{x^2} + x} \right)\left( {{x^2} + x - 2} \right) = 24\\
Dat:{x^2} + x = a\\
 \Leftrightarrow a\left( {a - 2} \right) = 24\\
 \Leftrightarrow {a^2} - 2a - 24 = 0\\
 \Leftrightarrow {a^2} + 4a - 6a - 24 = 0\\
 \Leftrightarrow \left( {a + 4} \right)\left( {a - 6} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
a + 4 = 0\\
a - 6 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} + x + 4 = 0\left( {vn} \right)\\
{x^2} + x - 6 = 0
\end{array} \right.\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 3} \right) = 0\\
 \Leftrightarrow x = 2;x =  - 3\\
Vậy\,x = 2;x =  - 3\\
b){x^4} + 2{x^3} - 2{x^2} + 2x - 3 = 0\\
 \Leftrightarrow {x^4} - {x^3} + 3{x^3} - 3{x^2} + {x^2} - x + 3x - 3 = 0\\
 \Leftrightarrow {x^3}\left( {x - 1} \right) + 3{x^2}\left( {x - 1} \right) + x\left( {x - 1} \right) + 3\left( {x - 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {{x^3} + 3{x^2} + x + 3} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 3} \right)\left( {{x^2} + 1} \right) = 0\\
 \Leftrightarrow x = 1;x =  - 3\\
Vậy\,x = 1;x =  - 3\\
c){x^3} - 6{x^2} + 12x + 19 = 0\\
 \Leftrightarrow {x^3} + {x^2} - 7{x^2} - 7x + 19x + 19 = 0\\
 \Leftrightarrow {x^2}\left( {x + 1} \right) - 7x\left( {x + 1} \right) + 19\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} - 7x + 19} \right) = 0\\
 \Leftrightarrow x + 1 = 0\\
 \Leftrightarrow x =  - 1\\
Vậy\,x =  - 1\\
d){x^3} + 5{x^2} - 4x - 20 = 0\\
 \Leftrightarrow {x^2}\left( {x + 5} \right) - 4\left( {x + 5} \right) = 0\\
 \Leftrightarrow \left( {x + 5} \right)\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow x =  - 5;x =  - 2;x = 2\\
Vậy\,x =  - 5;x =  - 2;x = 2\\
e){x^4} - 5{x^3} - 12{x^2} - 5x + 1 = 0\\
 \Leftrightarrow {x^4} + {x^3} - 6{x^3} - 6{x^2} - 6{x^2} - 6x + x + 1 = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {{x^3} - 6{x^2} - 6x + 1} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {{x^3} + {x^2} - 7{x^2} - 7x + x + 1} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x + 1} \right)\left( {{x^2} - 7x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
{x^2} - 2.x.\dfrac{7}{2} + \dfrac{{49}}{4} = \dfrac{{45}}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
{\left( {x - \dfrac{7}{2}} \right)^2} = {\left( {\dfrac{{3\sqrt 5 }}{2}} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
x = \dfrac{{7 \pm 3\sqrt 5 }}{2}
\end{array} \right.\\
Vậy\,x =  - 1;x = \dfrac{{7 \pm 3\sqrt 5 }}{2}
\end{array}$