Đáp án: $T = 1$
Giải thích các bước giải:
$\begin{array}{l}
4{x^2} - \dfrac{x}{2} = 1\\
\Leftrightarrow 4{x^2} - \dfrac{1}{2}.x - 1 = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{1}{2}:4 = \dfrac{1}{8}\\
{x_1}.{x_2} = - \dfrac{1}{4}
\end{array} \right.\\
T = {\left( {3{x_1} - 2} \right)^3}.{\left( {3{x_2} - 2} \right)^3}\\
= {\left[ {\left( {3{x_1} - 2} \right)\left( {3{x_2} - 2} \right)} \right]^3}\\
= {\left[ {\left( {9{x_1}{x_2} - 6\left( {{x_1} + {x_2}} \right) + 4} \right)} \right]^3}\\
= {\left( {9.\dfrac{{ - 1}}{4} - 6.\dfrac{1}{8} + 4} \right)^3}\\
= 1
\end{array}$
