Đáp án:
$\begin{array}{l}
a){x^4} - 2{x^3} + 10{x^2} - 20x = 0\\
\Rightarrow {x^3}\left( {x - 2} \right) + 10x\left( {x - 2} \right) = 0\\
\Rightarrow x.\left( {x - 2} \right)\left( {{x^2} + 10} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x - 2 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
\text{Vậy}\,x = 0;x = 2\\
b){\left( {2x - 3} \right)^2} = {\left( {x + 5} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
2x - 3 = x + 5\\
2x - 3 = - x - 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 8\\
3x = - 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 8\\
x = - \frac{2}{3}
\end{array} \right.\\
\text{Vậy}\,x = 8;x = - \frac{2}{3}\\
c){x^2}\left( {x - 1} \right) - 4{x^2} + 8x - 4 = 0\\
\Rightarrow {x^2}\left( {x - 1} \right) - 4\left( {{x^2} - 2x + 1} \right) = 0\\
\Rightarrow {x^2}\left( {x - 1} \right) - 4{\left( {x - 1} \right)^2} = 0\\
\Rightarrow \left( {x - 1} \right)\left( {{x^2} - 4\left( {x - 1} \right)} \right) = 0\\
\Rightarrow \left( {x - 1} \right)\left( {{x^2} - 4x + 4} \right) = 0\\
\Rightarrow \left( {x - 1} \right){\left( {x - 2} \right)^2} = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.\\
\text{Vậy}\,x = 1;x = 2
\end{array}$
