Question

Дам 23 балла!

Answer 1

  $8\cdot \frac{3^{x-2}}{3^{x}-2^{x}}>1+(\frac{2}{3})^{x}\\\\8\cdot \frac{3^{x}\cdot 3^{-2}}{3^{x}(1-(\frac{2}{3}^{x}))}>1+(\frac{2}{3})^{x}\\\\\frac{8\cdot \frac{1}{9}}{1-(\frac{2}{3})^{x}}-1-(\frac{2}{3})^{x}>0\\\\t=(\frac{2}{3})^{x}>0\; ,\; \; \frac{8}{9(1-t)}-1-t>0\\\\\frac{8-9+9t-9t+9t^2}{9(1-t)}>0\\\\\frac{9t^2-1}{9(1-t)}>0\; ,\; \frac{(3t-1)(3t+1)}{-9(t-1)}>0\\\\\frac{9(t-\frac{1}{3})(t+\frac{1}{3})}{9(t-1)}<0\\\\- - - - (-\frac{1}{3})+ + + + (\frac{1}{3})- - - (1)+ + + + +$

$t\in (-\infty ,-\frac{1}{3})U(\frac{1}{3},1)$

$(\frac{2}{3})^{x}<-\frac{1}{3}\; \; net\; reshenij\\\\\frac{1}{3}<(\frac{2}{3})^{x}<1\; \; \to \; \; (\frac{2}{3})^{log_{\frac{2}{3}}{\frac{1}{3}}}<(\frac{2}{3})^{x}<(\frac{2}{3})^0\\\\-log_{\frac{2}{3}}{3}<x<0$

Answer 2

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