Question

Решите неравенство. Подробнее пожалуйста. Пример в фото

Answer 1

$log_{0,5}(2x+1)<log_{2}(2-3x)\; \; \; ,\; \; \; ODZ:\; \left \{ {{2x+1>0} \atop {2-3x>0}} \right.\; \left \{ {{x>-\frac{1}{2}} \atop {x<\frac{2}{3}}} \right.\\\\\underline {x\in (-\frac{1}{2}\, ;\, \frac{2}{3})}\\\\-log_2(2x+1)<log_2(2-3x)\\\\a=2>1\; \; \; \Rightarrow \; \; \; (2x+1)^{-1}<2-3x\; ,$

$\frac{1}{2x+1}<2-3x\; \; ,\; \; \frac{1-(2x+1)(2-3x)}{2x+1}<0\; \; ,\; \frac{6x^2-x-1}{2x+1}<0\\\\ \frac{6(x-\frac{1}{2})(x+\frac{1}{3})}{2x+1}<0\\\\znaki\, :\; \; \; ---(-\frac{1}{2})+++(-\frac{1}{3})---(\frac{1}{2})+++\\\\x\in (-\infty \, ;\, -\frac{1}{2})\cup (-\frac{1}{3}\, ;\, \frac{1}{2})\\\\\left \{ {{x\in (-\frac{1}{2}\, ;\, \frac{2}{3})\qquad \qquad } \atop {x\in (-\infty \, ;\, -\frac{1}{2})\cup (-\frac{1}{3}\, ;\, \frac{1}{2})}} \right. \; \; \Rightarrow \\\\Otvet:\; \; x\in (-\frac{1}{3}\, ;\, \frac{1}{2})$