Question

Нужно сделать заменой 2 и 4

Answer 1

держи 2 и 4 решение)))))Удачи

Answer 2

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

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

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