Question

Номер 17, подробно, чтобы я смотрела сама понять и разобраться

Answer 1

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