Question

Сроочно,решите неравенство

Answer 1

вот ответ, если что спрашивай)

Answer 2

$\frac{2}{3^{x}-1}\leq \frac{7}{9^{x}-2}\; \; ,\; \; ODZ:\; \left \{ {{3^{x}\ne 1} \atop {9^{x}\ne 2}} \right. \; \left \{ {{x\ne 0} \atop {x\ne log_92}} \right. \\\\\frac{2(9^{x}-2)-7(3^{x}-1)}{(3^{x}-1)(9^{x}-2)}\leq 0\\\\\frac{2\cdot 9^{x}-4-7\cdot 3^{x}+7}{(3^{x}-1)((3^{x})^2-2)}\leq 0\; ,\; \; \frac{2\cdot 9^{x}-7\cdot 3^{x}+3}{(3^{x}-1)(3^{x}-\sqrt2)(3^{x}+\sqrt2)}\leq 0\\\\2\cdot (3^{x})^2-7\cdot 3^{x}+3=0\; \; \to \; \; 2t^2-7t+3=0\; ,\; \; t_1=\frac{1}{2}\; ,\; t_2=3\\\\3^{x}=\frac{1}{2}\; ,\; \; 3^{x}=3$

$\frac{2\cdot (t-\frac{1}{2})(t-3)}{(t-1)(t-\sqrt2)(t+\sqrt2)}\leq 0\; ,\; \; t=3^{x}>0\; ,\\\\znaki:\; \; ...(-\sqrt2)....(0)+++[\, \frac{1}{2}\, ]---(1)+++(\sqrt2)---[\, 3\, ]+++\\\\t\in [\,\frac{1}{2}\, ,1)\cup (\sqrt2,3\, ]\\\\\frac{1}{2}\leq 3^{x}<1\; \; \to \; \; log_3\frac{1}{2}\leq x<0\; \; ,\; \; -log_32\leq x<0\\\\\sqrt2<3^{x}\leq 3\; \; \to \; \; log_3\sqrt2<x\leq 1\; \; ,\; \; \frac{1}{2}log_32<x\leq 1\\\\x\in [-log_32,0\, )\cup (\, \frac{1}{2}log_32,1\, ]$

$\star \; \; \; b=a^{log_ab}\; \; \Rightarrow \; \; \frac{1}{2}=3^{log_3\frac{1}{2}}\; \; ,\; \; \sqrt2=3^{log_3\sqrt2}\; \; \star$