Giải thích các bước giải:
$\begin{array}{l}
1){x^3} - 4{x^2} + 4x = 0\\
\Leftrightarrow x\left( {{x^2} - 4x + 4} \right) = 0\\
\Leftrightarrow x{\left( {x - 2} \right)^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.
\end{array}$
Vậy $x \in \left\{ {0;2} \right\}$
$\begin{array}{l}
2)3x\left( {x - 2} \right) - x\left( {1 + 3x} \right) = 14\\
\Leftrightarrow 3{x^2} - 6x - x - 3{x^2} = 14\\
\Leftrightarrow - 7x = 14\\
\Leftrightarrow x = - 2
\end{array}$
Vậy $x \in \left\{ { - 2} \right\}$
$\begin{array}{l}
3){(2x - 1)^2} - (2x + 5)(2x + 3) = 32\\
\Leftrightarrow 4{x^2} - 4x + 1 - 4{x^2} - 16x - 15 = 32\\
\Leftrightarrow - 20x = 46\\
\Leftrightarrow x = \dfrac{{ - 23}}{{10}}
\end{array}$
Vậy $x \in \left\{ {\dfrac{{ - 23}}{{10}}} \right\}$
$\begin{array}{l}
4){(x + 2)^2} - (x - 2)(x + 2) = 16\\
\Leftrightarrow {x^2} + 4x + 4 - {x^2} + 4 = 16\\
\Leftrightarrow 4x = 8\\
\Leftrightarrow x = 2
\end{array}$
Vậy $x \in \left\{ 2 \right\}$
$\begin{array}{l}
5)(x + 2)({x^2} - 2x + 4) - x({x^2} + 2) = 1\\
\Leftrightarrow {x^3} + 8 - {x^3} - 2x - 1 = 0\\
\Leftrightarrow - 2x + 7 = 0\\
\Leftrightarrow x = \dfrac{{ - 7}}{2}
\end{array}$
Vậy $x \in \left\{ {\dfrac{{ - 7}}{2}} \right\}$
