Question

Дам 23 балла!

$8* frac{3^{x-2}}{3^{x}-2^{x}} > 1+ (frac{2}{3})^{x}$

Answer 1

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

$tin (-infty ,-frac{1}{3})U(frac{1}{3},1)$

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