Question

помогите пожалуйста​

Answer 1

$\left \{ {{9^{-x}-9^{-y}=-\frac{2}{9}} \atop {3^{x+y}=3\sqrt3}} \right.\; \left \{ {{3^{-2x}-3^{-2y}=-\frac{2}{9}} \atop {3^{x+y}=3^{3/2}} \right.\; \left \{ {{3^{-2x}-3^{-2y}=-\frac{2}{9}} \atop {x+y=3/2}} \right. \; \left \{ {{3^{-2x}-3^{2x-3}+\frac{2}{9}=0} \atop {y=3/2-x}} \right.\\\\\\a)\; \; \frac{1}{3^{2x}}-\frac{3^{2x}}{3^3}+\frac{2}{9}=0\\\\t=3^{2x}>0\; \; ,\; \; \frac{1}{t}-\frac{t}{27}+\frac{2}{9}=0\; ,\; \; \frac{27-t^2+6t}{27t}=0\; ,\\\\t^2-6t-27=0\; ,\; \; t_1=-3<0\; (!)\; ,\; t_2=9\; \; (teorema\; Vieta)$

$3^{2x}=9\; \; \to \; \; 3^{2x}=3^2\; \; \to \; \; 2x=2\; ,\; \; x=1\\\\b)\; \; y=\frac{3}{2}-x=\frac{3}{2}-1=\frac{1}{2}\\\\Otvet:\; \; (1\, ,\, \frac{1}{2})\; .$