b) ĐK $\dfrac{2}{3}\le x \le 7$
$(\sqrt{3x-2}-4) +(1-\sqrt{7-x}) +(3x^3-17x^2-8x+12)=0$
$\Leftrightarrow \dfrac{3(x-6)}{\sqrt{3x-2}+4}+\dfrac{x-6}{1+\sqrt{7-x}}+(x-6)(x+1)(3x-2)=0$
$\Leftrightarrow \left [ \begin{matrix} x=6 \\ \\ \dfrac{3}{\sqrt{3x-2}+4}+\dfrac{1}{1+\sqrt{7-x}}+(x+1)(3x-2)=0 \ (*)\end{matrix} \right.$
$(*) > 0 \forall x \in [\dfrac{2}{3};\ 7]$ vậy $(*)$ vô nghiệm