`~rai~`
\(4x^2-\dfrac{x}{2}=1\\\Leftrightarrow 4x^2-\dfrac{1}{2}x-1=0\\\text{Theo hệ thức Vi-et,ta có:}\\\begin{cases}x_1+x_2=\dfrac{1}{8}\\x_1x_2=-\dfrac{1}{4}\end{cases}\\T=(3x_1-2)^3(3x_2-2)^3\\\quad=[(3x_1-2)(3x_2-2)]^3\\\quad=(9x_1x_2-6x_1-6x_2+4)^3\\\quad=[9x_1x_2-6(x_1+x_2)+4]^3\\\quad=\left[9.\left(-\dfrac{1}{4}\right)-6.\dfrac{1}{8}+4\right]^3\\\quad=1.\\\text{Vậy T=1.}\)
$# Băng Cướp Cầu Vồng.$
