Question

помогите решить!если можно,то распишите подробно,пожалуйста

Answer 1

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

$znaki:; ; +++[, 1, ]---(2)+++(3)---[, frac{7}{2}, ]+++\\tin [, 1;2)cup (3;, frac{7}{2}, ]\\a); ; 1leq (x+1)sqrt3<2; ; ,; ; frac{1}{sqrt3}leq x+1<frac{2}{sqrt3}; ; ,; ; frac{sqrt3}{3}leq x+1<frac{2sqrt3}{3}; ,\\frac{sqrt3-3}{3}leq x<frac{2sqrt3-3}{3}\\b); ; 3<(x+1)sqrt3leq frac{7}{2}; ; ,; ; frac{3}{sqrt3}<x+1leq frac{7}{2sqrt3} ; ; ,; ; sqrt3<x+1leq frac{7sqrt3}{6}; ,\\sqrt3-1<xleq frac{7sqrt3-6}{6}$

$Otvet:; ; xin Big [, frac{xsqrt3-3}{3}, ;, frac{2sqrt3-3}{3}Big )cup Big (sqrt3-1, ;, frac{7sqrt3-6}{6}, Big ]$