Question

Решить неравенство .Заранее спасибо

Answer 1

$\sqrt{3^{1-x}-\frac{24}{1-3^{x}}}>\sqrt{9-\frac{8}{1-3^{x}}}\\\\ODZ:\; \; a)\; 3^{1-x}-\frac{24}{1-3^{x}}\geq 0\; ,\; \; \frac{3}{3^{x}}-\frac{24}{1-3^{x}}\geq 0\; ,\\\\\frac{3-3\cdot 3^{x}-24\cdot 3^{x}}{3^{x}\cdot (1-3^{x})}\geq 0\; ,\; \; \frac{3-27\cdot 3^{x}}{3^{x}(1-3^{x})}\geq 0\; ,\; \; \frac{27\cdot 3^{x}-3}{3^{x}\cdot (3^{x}-1)}\geq 0\; ,\; \; \frac{3(9\cdot 3^{x}-1)}{3^{x}(3^{x}-1)}\geq 0\\\\zamena\; t=3^{x}>0:\; \; \frac{3(9t-1)}{t(t-1)}\geq 0\; ,\\\\ ---(0)+++(\frac{1}{9})---(1)+++\\\\t\in (0,\frac{1}{9})\cup (1,+\infty )\; \; \Rightarrow \; \; \left [ {{0<3^{x}<3^{-2}} \atop {3^{x}>1}} \right. \; \left [ {{x<-2} \atop {x>0}} \right. \\\\x\in (-\infty ,-2)\cup (0,+\infty )$

$b)\; \; 9-\frac{8}{1-3^{x}}\geq 0\\\\t=3^{x}>0\; ,\; \; 9-\frac{8}{1-t}\geq 0\; ,\; \; \frac{9-9t-8}{1-t}\geq 0\; ,\; \; \frac{1-9t}{1-t}\geq 0\; ,\; \; \frac{9t-1}{t-1}\geq 0\; ,\\\\+++(\frac{1}{9})---(1)+++\\\\t\in (-\infty ,\frac{1}{9})\cup (1,+\infty )\; \to \; \; \left [ {{0<3^{x}<3^{-2}} \atop {3^{x}>1}} \right. \; ,\; \; \left [ {{x<-2} \atop {x>0}} \right. \\\\x\in (-\infty ,-2)\cup (0,+\infty )\\\\c)\; \; ODZ:\; \; x\in (-\infty ,-2)\cup (0,+\infty )$

$d)\; \; 3^{1-x}-\frac{24}{1-3^{x}}>9-\frac{8}{1-3^{x}}\\\\t=3^{x}>0:\; \; \frac{3}{t}-\frac{24}{1-t}>9-\frac{8}{1-t} \; ,\; \; \frac{3(1-t)-24t}{t\, (1-t)}>\frac{9(1-t)-8}{1-t}\; ,\\\\\frac{3-27t}{t\, (1-t)}-\frac{1-9t}{1-t}>0\; ,\\\\\frac{3-27t-t+9t^2}{t\, (1-t)}>0\; ,\; \; \frac{9t^2-28t+3}{t\, (1-t)}>0\; ,\; \; \frac{9\, (t-3)(t-\frac{1}{9})}{t\, (1-t)}>0\; ,\\\\\frac{9\, (t-3)(t-\frac{1}{9})}{t\, (t-1)}<0\\\\+++(0)---(\frac{1}{9})+++(1)---(3)+++\\\\t\in (0,\frac{1}{9})\cup (1,3)\; \; \to \; \; \left \{ {{0<3^{x}<3^{-2}} \atop {1<3^{x}<3}} \right.$

$\left [ {{x<-2} \atop {0<x<1}} \right. \; \; \Rightarrow \; \; x\in (-\infty ,-2)\cup (0,1)\\\\e)\; \; \left \{ {{x\in (-\infty ,-2)\cup (0,+\infty )} \atop {x\in (-\infty ,-2)\cup (0,1)}} \right. \; \; \Rightarrow \; \; \underline {x\in (-\infty ,-2)\cup (0,1)}$