Đáp án:
$\begin{array}{l}
a)3{x^3} + 6{x^2} - 4x = 0\\
\Leftrightarrow x.\left( {3{x^2} + 6x - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
3{x^2} + 6x - 4 = 0\left( * \right)
\end{array} \right.\\
{\Delta _*} = {3^2} - 3.\left( { - 4} \right) = 21\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{{ - 3 + \sqrt {21} }}{3}\\
x = \dfrac{{ - 3 - \sqrt {21} }}{3}
\end{array} \right.\\
b){\left( {{x^2} + x + 1} \right)^2} = {\left( {4x - 1} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + x + 1 = 4x - 1\\
{x^2} + x + 1 = - 4x + 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 3x + 2 = 0\\
{x^2} + 5x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {x - 1} \right)\left( {x - 2} \right) = 0\\
x\left( {x + 5} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = 0\\
x = - 5
\end{array} \right.\\
c){x^3} - 5{x^2} - x + 5 = 0\\
\Leftrightarrow {x^2}\left( {x - 5} \right) - \left( {x - 5} \right) = 0\\
\Leftrightarrow \left( {x - 5} \right)\left( {{x^2} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
{x^2} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 1\\
x = - 1
\end{array} \right.\\
d){x^2} + 2\sqrt 2 x + 4 = 3\left( {x + \sqrt 2 } \right)\\
\Leftrightarrow {x^2} + 2\sqrt 2 .x - 3x + 4 - 3\sqrt 2 = 0\\
\Leftrightarrow {x^2} + \left( {2\sqrt 2 - 3} \right).x + 4 - 3\sqrt 2 = 0\\
\Leftrightarrow \Delta = {\left( {2\sqrt 2 - 3} \right)^2} - 4.\left( {4 - 3\sqrt 2 } \right)\\
= 8 - 2.2\sqrt 2 .3 + 9 - 16 + 12\sqrt 2 \\
= 1\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{3 - 2\sqrt 2 + \sqrt 1 }}{1} = 4 - 2\sqrt 2 \\
x = \dfrac{{3 - 2\sqrt 2 - 1}}{1} = 2 - 2\sqrt 2
\end{array} \right.
\end{array}$
