Question

Помогите, пожалуйста
Даю 50 баллов

Answer 1

$2)\; \; 3^{x}+\Big (3\, \sqrt{\sqrt{10}-1}\Big )^{x}=2\, \Big (\sqrt{\sqrt{10}+1}\Big )^{x}\\\\\star \; \; (\sqrt{\sqrt{10}-1})^{x}\cdot (\sqrt{\sqrt{10}+1})^{x}=\Big (\sqrt{(\sqrt{10}-1)(\sqrt{10}+1)}\Big )^{x}=\\\\=(\sqrt{10-1})^{x}=(\sqrt9)^{x}=3^{x}\; \; \Rightarrow \; \; (\sqrt{\sqrt{10}+1})^{x}=\frac{3^{x}}{(\sqrt{\sqrt{10}-1})^{x}}\; \; \star \\\\t=(\sqrt{\sqrt{10}-1})^{x}>0\; \; ,\\\\3^{x}+3^{x}\cdot t=2\cdot \frac{3^{x}}{t}\; |:3^{x}\; ,\; (3^{x}>0)\\\\1+t=\frac{2}{t}\; \; ,\; \; 1+t-\frac{2}{t}=0\; \; ,\; \; \frac{t^2+t-2}{t}=0\; ,\; t>0$

$t^2+t-2=0\; \; ,\; \; t_1=-2<0\; \; ,\; \; t_2=1>0\\\\(\sqrt{\sqrt{10}-1})^{x}=1\; \; ,\; \; (\sqrt{\sqrt{10}-1})^{x}=(\sqrt{\sqrt{10}-1})^0\; \; \Rightarrow \; \; \boxed{x=0}$

$3)\; \; 4\cdot (\sqrt5+1)^{\frac{6x-3}{x-2}}\geq \frac{(\sqrt5-1)^{2x+1}}{16^2}\; \; ,\; \; \; ODZ:\; \; x-2\ne 0\; \to \; \; x\ne 2\\\\\star \; \; (\sqrt5-1)\cdot (\sqrt5+1)=5-1=4\; \; \Rightarrow \; \; (\sqrt5-1)=\frac{4}{\sqrt5+1}\star \\\\4\cdot (\sqrt5+1)^{\frac{6x-3}{x-2}}\geq \frac{(\frac{4}{\sqrt5+1})^{2x+1}}{4^{2x}}\\\\4\cdot (\sqrt5+1)^{\frac{6x-3}{x-2}}\geq \frac{4^{2x+1}}{4^{2x}\cdot (\sqrt5+1)^{2x+1}}\\\\4\cdot (\sqrt5+1)^{\frac{6x-3}{x-2}}-\frac{4}{(\sqty5+1)^{2x+1}} \geq 0\, |:4$

$\frac{(\sqrt5+1)^{\frac{6x-3}{x-2}}\cdot (\sqrt5+1)^{2x+1}-1}{(\sqrt5+1)^{2x+1}}\geq 0\; \; ,\; \; \; (\sqrt5+1)^{2x+1}>0\\\\(\sqrt5+1)^{\frac{6x-3}{x-2}+(2x+1)}-1\geq 0\\\\\star \; \; \frac{6x-3}{x-2}+2x+1=\frac{6x-3+2x^2+x-4x-2}{x-2}=\frac{2x^2+3x-5}{x-2}\; \; \star \\\\(\sqrt5+1)^{\frac{2x^2+3x-5}{x-2}}\geq 1\\\\(\sqrt5+1)^{\frac{2x^2+3x-5}{x-2}}\geq (\sqrt5+1)^0\\\\\frac{2x^2+3x-5}{x-2}\geq 0\; \; \; \; [\; 2x^2+3x-5=0\; ,\; \; x_1=-2,5\; \; ,\; \; x-2=1\; ]\\\\\frac{2\, (x+2,5)(x-1)}{x-2}\geq 0$

$znaki:\; \; \; ---[-2,5\, ]+++(2)---[\, 1\, ]+++\\\\\underline {\; x\in [-2,5\, ;\, 2)\cup [1\, ;+\infty )}$