Question

Answer 1

=>8x^2-x-2=0

x1+x2=1/8; x1x2=-1/4

\(T=\left[\left(3x_1-2\right)\left(3x_2-2\right)\right]^3\)

\(=\left[9x_1x_2-6x_1-6x_2+4\right]^3\)

\(=\left[9\cdot\dfrac{-1}{4}-6\cdot\dfrac{1}{8}+4\right]^3=1\)

Answer 2

\(\Leftrightarrow4x^2-\dfrac{x}{2}-1=0\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{1}{8}\\x_1x_2=-\dfrac{1}{4}\end{matrix}\right.\)

\(T=\left[\left(3x_1-2\right)\left(3x_2-2\right)\right]^3\)

\(=\left(9x_1x_2-6\left(x_1+x_2\right)+4\right)^3\)

\(=\left(-\dfrac{9}{4}-\dfrac{6}{8}+4\right)^3=1\)