Решить неравенство ㏒1-2x(x)< =3-㏒x() где x и 1/x-2 - основания логарифма

Ответ:

kamilaCOOL777

05.10.2020 17:38

$log_{x}(1-2x) \leq 3-log_{\frac{1}{x}-2}\; x\\\\ODZ:\; \; \{\; x\ \textgreater \ 0\; ,\; x\ne 1\; ,\; 1-2x\ \textgreater \ 0\; ,\frac{1}{x}-2\ \textgreater \ 0\; ,\; \frac{1}{x}-2\ne 1\; \}\\\\\{\; x\ \textgreater \ 0\; ,\; x\ne 1\; ,\; x\ \textless \ \frac{1}{2}\; ,\; 0\ \textless \ x\ \textless \ \frac{1}{2}\; ,\; x\ne \frac{1}{3}\; \}\; \to \; x\in (0,\frac{1}{3})\cup (\frac{1}{3},\frac{1}{2})\\\\\\log_{x}(1-2x) \leq 3-\frac{1}{log_{x}\, (\frac{1}{x}-2)}\\\\log_{x}(1-2x)+ \frac{1}{log_{x}\frac{1-2x}{x}} \leq 3$

$log_{x}(1-2x)+ \frac{1}{log_{x}(1-2x)-log_{x}x} \leq 3\\\\t=log_{x}(1-2x)\; \; \Rightarrow \; \; \; t+\frac{1}{t-1} \leq 3$

$t+\frac{1}{t-1}-3 \leq 0\; ,\; \; \; \frac{t^2-t+1-3(t-1)}{t-1} \leq 0\; ,\; \; \frac{t^2-4t+4}{t-1} \leq 0\\\\ \frac{(t-2)^2}{t-1} \leq 0\; \; \; ---(1)+++[2]++++\\\\t\ \textless \ 1\; \; \to \; \; \; log_{x}(1-2x)\ \textless \ 1\; ,\\\\log_{x}(1-2x)\ \textless \ log_{x}\, x\; \; \Leftrightarrow \; \; \; (x-1)(1-2x-x)\ \textless \ 0\\\\(x-1)(1-3x)\ \textless \ 0\; ,\; \; (x-1)(3x-1)\ \textgreater \ 0\\\\+++(\frac{1}{3})---(1)+++\\\\x\in (-\infty ,\frac{1}{3})\cup (1,+\infty )\\\\ \left \{ {{x\in (0,\frac{1}{3})\cup (\frac{1}{3},\frac{1}{2})} \atop {x\in (-\infty ,\frac{1}{3})\cup (1,+\infty )} \right.$

$x\in (0,\frac{1}{3})\; .$