Решите пожалуйста уравнение и неравенство

$\frac{x-6}{x^2-4x+4} <0\\\\\frac{x-6}{(x-2)^2}<0\\\\(x-2)^2 \geq 0\; pri\; x\in (-\infty;+\infty)\; \; \to \\\\x-6<0\\\\x<6\\\\x\in (-\infty;6)$

$25^{x+1}-34*15^{x}+9^{x+1}=0\\\\25*(5^{x})^2-34*5^{x}*3^{x}+9*(3^{x})^2=0|:(3^{x})^2\\\\25*(\frac{5}{3})^{2x}-34*(\frac{5}{3})^{x}+9=0$

0,\\\\25t^2-34t+9=0\\\\D/4=64,t_1=\frac{17-8}{25}=\frac{9}{25}=(\frac{3}{5})^2;t_2=1\\\\(\frac{5}{3})^{x}=(\frac{5}{3})^{-2},x=-2,\\\\(\frac{5}{3})^{x}=1,x=0" alt="t=(\frac{5}{3})^{x}>0,\\\\25t^2-34t+9=0\\\\D/4=64,t_1=\frac{17-8}{25}=\frac{9}{25}=(\frac{3}{5})^2;t_2=1\\\\(\frac{5}{3})^{x}=(\frac{5}{3})^{-2},x=-2,\\\\(\frac{5}{3})^{x}=1,x=0" align="absmiddle" class="latex-formula">