Question

решите пожалуйста 15 задание. ........

Answer 1

$3^{x}+8\cdot 3^{-x} \geq 9\\\\3^{x}=t\ \textgreater \ 0,\; \; t+\frac{8}{t}-9 \geq 0\; \; \to \frac{t^2-9t+8}{t} \geq 0\\\\ \frac{(t-1)(t-8)}{t} \geq 0\\\\---(0)+++[1]---[8]+++\; \; \; \; \left \{ {{0\ \textless \ t \leq 1} \atop {t \geq 8}} \right. \\\\3^{x} \leq 1\; ,\; 3^{x} \leq 3^0\; \; \to \; \; x \leq 0\\\\3^{x} \geq 8\; ,\; 3^{x} \geq 3^{log_38}\; \; \to \; \; x \geq log_38\\\\x\in (-\infty ;0\, ]\cup [\, log_38;+\infty )$