Решите неравенство (4-3x)^2 (2x+3)≤0.

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

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

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

\[x = \frac{4}{3} = 1\frac{1}{3};\ \ \ x = - 1,5\]

\[x \in ( - \infty; - 1,5\rbrack \cup \left\{ 1\frac{1}{3} \right\}.\]