Question

помогите решить!если можно,то распишите подробно,пожалуйста

Answer 1

$\frac{3}{2-(x+1)\sqrt3}+\frac{(x+1)\sqrt3-1}{(x+1)\sqrt3-3}\geq 3\\\\t=(x+1)\sqrt3\; \; ,\; \; \frac{3}{2-t}+\frac{t-1}{t-3}\geq 3\; \; ,\; \; ODZ:\; \; t\ne 3\; ,\; t\ne 2\\\\\frac{3t-9+(t-1)(2-t)}{(2-t)(t-3)}-3\geq 0\; \; ,\; \frac{3t-9+2t-t^2-2+t}{(2-t)(t-3)}-3\geq 0\\\\\frac{-t^2+6t-11-3(-t^2+5t-6)}{(2-t)(t-3)}\geq 0\; \; ,\; \; \frac{2t^2-9t+7}{-(t-2)(t-3)}\geq 0\\\\2t^2-9t+7=0\; ,\; \; D=25\; ,\; \; x_1=1\; ,\; \; x_2=\frac{7}{2}=3,5\\\\\frac{2(t-1)(t-\frac{7}{2})}{(t-2)(t-3)}\leq 0$

$znaki:\; \; +++[\, 1\, ]---(2)+++(3)---[\, \frac{7}{2}\, ]+++\\\\t\in [\, 1;2)\cup (3;\, \frac{7}{2}\, ]\\\\a)\; \; 1\leq (x+1)\sqrt3<2\; \; ,\; \; \frac{1}{\sqrt3}\leq x+1<\frac{2}{\sqrt3}\; \; ,\; \; \frac{\sqrt3}{3}\leq x+1<\frac{2\sqrt3}{3}\; ,\\\\\frac{\sqrt3-3}{3}\leq x<\frac{2\sqrt3-3}{3}\\\\b)\; \; 3<(x+1)\sqrt3\leq \frac{7}{2}\; \; ,\; \; \frac{3}{\sqrt3}<x+1\leq \frac{7}{2\sqrt3} \; \; ,\; \; \sqrt3<x+1\leq \frac{7\sqrt3}{6}\; ,\\\\\sqrt3-1<x\leq \frac{7\sqrt3-6}{6}$

$Otvet:\; \; x\in \Big [\, \frac{x\sqrt3-3}{3}\, ;\, \frac{2\sqrt3-3}{3}\Big )\cup \Big (\sqrt3-1\, ;\, \frac{7\sqrt3-6}{6}\, \Big ]$