Đáp án: ${x = \frac{1}{3}\,hoặc\,x = \frac{{ - 5}}{4}}$
Giải thích các bước giải:
$\begin{array}{l}
\left( {4x + 1} \right)\left( {12x - 1} \right)\left( {3x + 2} \right)\left( {x + 1} \right) = 28\\
\Rightarrow \left[ {\left( {4x + 1} \right)\left( {3x + 2} \right)} \right]\left[ {\left( {12x - 1} \right)\left( {x + 1} \right)} \right] = 28\\
\Rightarrow \left( {12{x^2} + 11x + 2} \right)\left( {12{x^2} + 11x - 1} \right) = 28\\
Đặt:12{x^2} + 11x = a\\
\Rightarrow \left( {a + 2} \right)\left( {a - 1} \right) = 28\\
\Rightarrow {a^2} + a - 2 = 28\\
\Rightarrow {a^2} + a - 30 = 0\\
\Rightarrow {a^2} + 6a - 5a - 30 = 0\\
\Rightarrow \left( {a + 6} \right)\left( {a - 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a + 6 = 0\\
a - 5 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
12{x^2} + 11x + 6 = 0\left( {vn} \right)\\
12{x^2} + 11x - 5 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{1}{3}\\
x = \frac{{ - 5}}{4}
\end{array} \right.
\end{array}$
