Пожалуйста помогите решить неравенство!

Применим метод рационализации.

$log_{0,25x^2}(\frac{x+12}{4}) \leq 1\; ;\; \; ODZ:\; \left \{ {{\frac{x+12}{4}\ \textgreater \ 0} \atop {0,25x^2\ \textgreater \ 0,\; \; 0,25x^2\ne 1}} \right. \; \left \{ {{x\ \textgreater \ -12} \atop {x\ne \pm 2}} \right. \\\\(\frac{x+12}{4}-0,25x^2)(0,25x^2-1) \leq 0\\\\(\frac{x+12}{4}-\frac{x^2}{4})(\frac{x^2}{4}-1) \leq 0\\\\\frac{x+12-x^2}{4}\cdot \frac{x^2-4}{4} \leq 0\\\\\frac{x^2-x-12}{4}\cdot \frac{(x-2)(x+2)}{4} \geq 0\\\\(x-4)(x+3)(x-2)(x+2) \geq 0\\\\+++(-3)---(-2)+++(2)---(4)+++\\\\x\in (-12 ,-3\, ]\cup (-2,2)\cup [\, 4,+\infty )$