Đáp án:

\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = 0\\
x = \sqrt 5  - 3\\
x =  - \sqrt 5  - 3
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = \dfrac{3}{4}\\
x = \dfrac{3}{7}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
{x^3} + 6{x^2} + 4x = 0\\
 \Leftrightarrow x\left( {{x^2} + 6x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} + 6x + 4 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} + 6x + 9 = 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} + 2.x.3 + {3^2} = 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{\left( {x + 3} \right)^2} = 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 3 = \sqrt 5 \\
x + 3 =  - \sqrt 5 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt 5  - 3\\
x =  - \sqrt 5  - 3
\end{array} \right.\\
b,\\
{\left( {4x - 3} \right)^2} - 3x.\left( {3 - 4x} \right) = 0\\
 \Leftrightarrow {\left( {4x - 3} \right)^2} - 3x.\left[ { - \left( {4x - 3} \right)} \right] = 0\\
 \Leftrightarrow {\left( {4x - 3} \right)^2} + 3x.\left( {4x - 3} \right) = 0\\
 \Leftrightarrow \left( {4x - 3} \right).\left[ {\left( {4x - 3} \right) + 3x} \right] = 0\\
 \Leftrightarrow \left( {4x - 3} \right)\left( {7x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
4x - 3 = 0\\
7x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{4}\\
x = \dfrac{3}{7}
\end{array} \right.
\end{array}\)