Question

Срочно! 3-й. Сколько сможете

Answer 1

$\#1).\; \; \; 3\cdot 9^{x}+6^{x}=2\cdot 4^{x}\; \Big |:(4^{x}=2^{2x})\\\\3\cdot (\frac{3}{2})^{2x}+(\frac{3}{2})^{x}-2=0\\\\t=(\frac{3}{2})^{x}>0\; \; \to \; \; \; 3t^2+t-2=0\; ,\; \; D=1+24=25\; ,\\\\t_1= \frac{-1-5}{6}=-1<0\; \; ,\; \; t_1=\frac{-1+5}{6}=\frac{2}{3}>0\\\\(\frac{3}{2})^{x}=\frac{2}{3}\; \; \to \; \; \; (\frac{3}{2})^{x}=(\frac{3}{2})^{-1}\; \; ,\; \; \boxed {x=-1}$

$\#2).\quad (2+\sqrt3)^{x}+(2-\sqrt3)^{x}>4\\\\\star \; \; (2+\sqrt3)(2-\sqrt3)=4-3=1\; \; \to \; \; (2-\sqrt3)=\frac{1}{2+\sqrt3}\; \; \star \\\\t=(2+\sqrt3)^{x}>0\; \; \to \; \; (2-\sqrt3)^{x}=\frac{1}{t}\; ,\\\\t+\frac{1}{t}-4>0\; \; \to \; \; \frac{t^2-4t+1}{t}>0\; ,\\\\t^2-4t+1=0\; \to \; D/4=4-1=3\; ,\; \; t_1=2-\sqrt3\; ,\; \; t_2=2+\sqrt3\\\\a)\; (2+\sqrt3)^{x}=2-\sqrt3\; \; \to \; \; (2+\sqrt3)^{x}=\frac{1}{2+\sqrt3}=(2+\sqrt3)^{-1}\\\\x=-1\\\\b)\; \; (2+\sqrt3)^{x}=2+\sqrt3\; \; \to \; \; x=1\\\\Otvet:\; \; x=-1\; ,\; \; x=1\; .$