Решить уравнение:

Метод рационализации:

$log_{12x^2-5x-2} (4x+1)\leq 0\\\\ODZ:\; \left \{ {{12x^2-5x-2\ne 1 ,\; 12x^2-5x-2\ \textgreater \ 0} \atop {4x+1\ \textgreater \ 0}} \right. \; \left \{ {{x\ne \frac{3}{4},x\ne -\frac{1}{3},12x^2-5x-2\ \textgreater \ 0} \atop {x\ \textgreater \ -\frac{1}{4}}} \right.$

$12x^2-5x-2\ \textgreater \ 0\; ,\; D=25+96=121,\\\\x_1= \frac{5-11}{24} =-\frac{1}{4}\; ,\; x_2= \frac{5+11}{24} =\frac{2}{3}\\\\12(x+\frac{1}{4})(x-\frac{2}{3})\ \textgreater \ 0\; ,\; \; +++(-\frac{1}{4})---(\frac{2}{3})+++\\\\x\in (-\infty ,-\frac{1}{4})\cup (\frac{2}{3},+\infty )\; \; \Rightarrow\\\\ODZ:\; x\in ( \frac{2}{3}, \frac{3}{4})\cup ( \frac{3}{4},+\infty )$

$\star \quad log_{h}f \leq 0\; \; \Leftrightarrow \; \; (h-1)(f-1) \leq 0\\\\(12x^2-5x-2-1)(4x+1-1) \leq 0\\\\(12x^2-5x-3)\cdot 4x \leq 0\\\\12\cdot (x-\frac{3}{4})(x+\frac{1}{3})\cdot 4x \leq 0\\\\Znaki:\; \; ---[-\frac{1}{3}]+++[\, 0\, ]---[\frac{3}{4}]+++$

$x\in (-\infty ,-\frac{1}{3}]\cup [\, 0,\frac{3}{4}\, ]$

$Otvet:x\in (\frac{2}{3},\frac{3}{4})$