Question

Решить неравенство
$(\frac{1}{\sqrt{10}-3})^{\frac{x+2}{1-x}}<(\frac{1}{\sqrt{10}+3})^{\frac{2x+8}{x+1}}$

Answer 1

$(\frac{1}{\sqrt{10}-3})^{\frac{x+2}{1-x}}<(\frac{1}{\sqrt{10}+3})^{\frac{2x+8}{x+1}}\; ,\; \; \; \; ODZ:\; \; x\ne \pm 1\\\\ (\sqrt{10}+3)(\sqrt{10}-3)=10-9=1\; \; \Rightarrow \; \; (\sqrt{10}-3)=\frac{1}{\sqrt{10}+3}\; ,\\\\a=\sqrt{10}+3>1\; \; ,\; \; \sqrt{10}-3=\frac{1}{a}\; \; ,\; \; \; \; a^{\frac{x+2}{1-x}}<(\frac{1}{a})^{\frac{2x+8}{x+1}}\; \; ,\\\\a^{\frac{x+2}{1-x}}<a^{-\frac{2x+8}{x+1}}$

$a>1\; \; \Rightarrow \; \; \; \frac{x+2}{1-x}<-\frac{2x+8}{x+1}\; \; ,\; \; \frac{x+2}{-(x-1)}+\frac{2x+8}{x+1}<0\; \; ,\; \; \frac{2x+8}{x+1}-\frac{x+2}{x-1}<0\; ,\\\\\frac{(2x+8)(x-1)-(x+2)(x+1)}{(x-1)(x+1)}<0\; ,\; \; \frac{2x^2+6x-8-(x^2+3x+2)}{(x-1)(x+1)}<0\; ,\\\\\frac{x^2+3x-10}{(x-1)(x+1)}<0\; ,\; \; \frac{(x-2)(x+5)}{(x-1)(x+1)}<0\; \; \; \; (x\ne \pm 1)\\\\znaki:\; \; \; +++(-5)---(-1)+++(1)---(2)+++\\\\\underline {x\in (-5,-1)\cup (1,2)}\\\\celue\; \; resheniya:\; \; x=-4\; ,\; -3\; ,\; -2\; .\\\\-4-3-2=-9$