0\\\\(\frac{5^{x}}{2^{x}})^3+5\cdot (\frac{5^{x}}{2^{x}})^2-\frac{5^{x}}{2^{x}}-5=0\\\\\underline {t=(\frac{5}{2})^{x}>0}\; \; \to \; \; \; t^3+5t^2-t-5=0\; ,\\\\t^2\, (t+5)-(t+5)=0\\\\(t+5)\cdot (t^2-1)=0\\\\(t+5)\cdot (t-1)\cdot (t+1)=0\\\\t_1=-5\; ,\; t_2=1\; ,\; t_3=-1\\\\\underline {t>0\; (!!!)}\; \; \to \; \; t_2=1>0\; \; podxodit\\\\(\frac{5}{2})^{x}=1\; \; \Rightarrow \; \; (\frac{5}{2})^{x}=(\frac{5}{2})^0\; \; \Rightarrow \; \; \boxed {\; x=0\; }" alt="5^{3x}+5^{2x+1}\cdot 2^{x}-20^{x}-5\cdot 8^{x}=0\\\\(5^{x})^3+(5^{x})^2\cdot 5\cdot 2^{x}-(2^{x})^2\cdot 5^{x}-5\cdot (2^{x})^3=0\, |:(2^{x})^3>0\\\\(\frac{5^{x}}{2^{x}})^3+5\cdot (\frac{5^{x}}{2^{x}})^2-\frac{5^{x}}{2^{x}}-5=0\\\\\underline {t=(\frac{5}{2})^{x}>0}\; \; \to \; \; \; t^3+5t^2-t-5=0\; ,\\\\t^2\, (t+5)-(t+5)=0\\\\(t+5)\cdot (t^2-1)=0\\\\(t+5)\cdot (t-1)\cdot (t+1)=0\\\\t_1=-5\; ,\; t_2=1\; ,\; t_3=-1\\\\\underline {t>0\; (!!!)}\; \; \to \; \; t_2=1>0\; \; podxodit\\\\(\frac{5}{2})^{x}=1\; \; \Rightarrow \; \; (\frac{5}{2})^{x}=(\frac{5}{2})^0\; \; \Rightarrow \; \; \boxed {\; x=0\; }" align="absmiddle" class="latex-formula">