Đáp án:

\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = 0\\
x = 4\\
x =  - 4
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
c,\\
\left[ \begin{array}{l}
x = 8\\
x =  - \dfrac{2}{3}
\end{array} \right.\\
d,\\
\left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a,\\
{x^3} - 16x = 0\\
 \Leftrightarrow x.\left( {{x^2} - 16} \right) = 0\\
 \Leftrightarrow x.\left( {{x^2} - {4^2}} \right) = 0\\
 \Leftrightarrow x\left( {x - 4} \right)\left( {x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 4 = 0\\
x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 4\\
x =  - 4
\end{array} \right.\\
b,\\
{x^4} - 2{x^3} + 10{x^2} - 20x = 0\\
 \Leftrightarrow x.\left( {{x^3} - 2{x^2} + 10x - 20} \right) = 0\\
 \Leftrightarrow x.\left[ {\left( {{x^3} - 2{x^2}} \right) + \left( {10x - 20} \right)} \right] = 0\\
 \Leftrightarrow x.\left[ {{x^2}.\left( {x - 2} \right) + 10\left( {x - 2} \right)} \right] = 0\\
 \Leftrightarrow x.\left( {x - 2} \right).\left( {{x^2} + 10} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 2 = 0\\
{x^2} + 10 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
{x^2} =  - 10\,\,\,\left( L \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
c,\\
{\left( {2x - 3} \right)^2} = {\left( {x + 5} \right)^2}\\
 \Leftrightarrow {\left( {2x - 3} \right)^2} - {\left( {x + 5} \right)^2} = 0\\
 \Leftrightarrow \left[ {\left( {2x - 3} \right) - \left( {x + 5} \right)} \right].\left[ {\left( {2x - 3} \right) + \left( {x + 5} \right)} \right] = 0\\
 \Leftrightarrow \left( {2x - 3 - x - 5} \right)\left( {2x - 3 + x + 5} \right) = 0\\
 \Leftrightarrow \left( {x - 8} \right)\left( {3x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 8 = 0\\
3x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x =  - \dfrac{2}{3}
\end{array} \right.\\
d,\\
{x^2}\left( {x - 1} \right) - 4{x^2} + 8x - 4 = 0\\
 \Leftrightarrow {x^2}\left( {x - 1} \right) - \left( {4{x^2} - 8x + 4} \right) = 0\\
 \Leftrightarrow {x^2}\left( {x - 1} \right) - 4.\left( {{x^2} - 2x + 1} \right) = 0\\
 \Leftrightarrow {x^2}.\left( {x - 1} \right) - 4.\left( {{x^2} - 2.x.1 + {1^2}} \right) = 0\\
 \Leftrightarrow {x^2}.\left( {x - 1} \right) - 4.{\left( {x - 1} \right)^2} = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left[ {{x^2} - 4.\left( {x - 1} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left( {{x^2} - 4x + 4} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left( {{x^2} - 2.x.2 + {2^2}} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right).{\left( {x - 2} \right)^2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
{\left( {x - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.
\end{array}\)