Нужно сделать заменой 2 и 4

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Автор - spirit95

держи 2 и 4 решение)))))Удачи

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Проверено экспертом

Автор - NNNLLL54

$2); ; frac{7}{(x^2+x+1)^2}leq frac{8}{x^2+x+1}-1\\t=x^2+x+1>0; ; ,; ; frac{7}{t^2}-frac{8}{t}+1leq 0; ; ,; ; frac{t^2-8t+7}{t^2} leq 0; ; ,\\frac{(t-1)(t-7)}{t^2}leq 0; ; ,; ; znaki:; ; +++(0)+++[, 1, ]---[, 7, ]+++\\tin [, 1,7, ]; ; Rightarrow ; ; ; 1leq x^2+x+1leq 7\\left { {{x^2+x+1leq 7} atop {x^2+x+1geq 1}} right. ; left { {{x^2+x-6leq 0} atop {x^2+xgeq 0}} right. ; left { {{(x-2)(x+3)leq 0} atop {x(x+1)geq 0}} right. ; left { {{xin [-3,2, ]} atop {xin (-infty ,-1, ]cup [, 0,+infty )}} right.$

$xin [-3,-1, ]cup [, 0,2, ]\\\4); ; frac{2}{|x+3|-4}leq frac{1}{|x+3|-2}; ; ,; ; ; ODZ:; left { {{|x+3|-4ne 0} atop {|x+3|-2ne 0}} right. \\t=|x+3|geq 0; ; ,; ; ; frac{2}{t-4}-frac{1}{t-2}leq 0; ; ,; ; frac{t}{(t-4)(t-2)}leq 0; ; ,; ; tne 2; ,; tne 4\\znaki:; ; ---[, 0, ]+++(2)---(4)+++\\tin (-infty ,0, ]cup (2,4); ; Rightarrow ; ; left [ {{|x+3|leq 0} atop {left { {{|x+3|<4} atop {|x+3|>2}} right. }} right.$

$left [ {{x+3=0} atop {left { {{-4<x+3<4} atop {x+3>2; ; ili; ; x+3<-2}} right. }} right. ; ; left [ {{x=-3} atop {left { {{-7<x<1} atop {x>1; ; ili; ; x<-5}} right. }} right.; ; left [ {{x=-3} atop {left { {{xin (-7,1)} atop {xin (-infty ,-5)cup (1,+infty )}} right. }} right. \\\left [ {{x=-3} atop {xin (-7,-5)cup (-1,1)}} right. ; ; Rightarrow ; ; xin (-7,-5)cup {-3}cup (-1,1)$