Question

ПОМОГИТЕ РЕШИТЬ!!! ХОТЯ БЫ ПРЕОБРАЗОВАТЬ!

Answer 1

$log_{\frac{x}{x-1}}5 \leq log_{\frac{x}{2}}5\; ;\; \; ODZ:\; \left \{ {{\frac{x}{x-1}>0,\frac{x}{x-1}\ne 1} \atop {\frac{x}{2}>0,\frac{x}{2}\ne 1}} \right. \\\\\frac{1}{log_5\frac{x}{x-1}} \leq \frac{1}{log_5\frac{x}{2}}\; \; \to \; \; log_5\frac{x}{x-1} \geq log_5\frac{x}{2}\; \; \to \; \; \frac{x}{x-1} \geq \frac{x}{2}\\\\\frac{2x-x^2+x}{2(x-1)} \geq 0\\\\\frac{x^2-3x}{2(x-1)} \leq 0\\\\\frac{x(x-3)}{2(x-1)}\leq 0 \\\\--0++1--3++\\\\x\in(-\infty,0\, ]U(1,3\, ] \\\\Otvet:\; x\in (1,2)U(2,3\ ]$

ОДЗ: хЄ(1,2)U(2,+беск)