Решите неравенство (3-4x)^2 (3x+2)≤0.

\[(3 - 4x)^{2}(3x + 2) \leq 0\]

\[(3x + 2)(3 - 4x)(3 - 4x) \leq 0\]

\[x = \frac{3}{4};\ \ x = - \frac{2}{3}\]

\[Ответ:x \in \left( - \infty; - \frac{2}{3} \right\rbrack \cup \left\{ 0,75 \right\}.\]