Question

Кто может решить, не проходите мимо
УРАВНЕНИЕ:

Answer 1

Пошаговое объяснение:

$\sqrt{5x^2+24x+28}-\sqrt[3]{5x^2+24x+27}\leq 1 .$

ОДЗ:

$5x^2+24x+28\geq 0\\5x^2+10x+14x+28\geq 0\\5x*(x+2)+14*(x+2)\geq 0\\(x+2)*(5x+14)\geq 0.$

-∞__+__-2,8__-__-2__+__+∞

x∈(-∞;-2,8]U[-2;+∞).

Пусть: $\sqrt[6]{5x^2+24x+28} =t\geq 0\ \ \ \ \ \Rightarrow\\$

$t^3-(t-1)^2\leq 1\\t^3-(t^2-2t+1)-1\leq 0\\t^3-t^2+2t-1-1\leq 0\\t^3-t^2+2t-2\leq 0\\t^2*(t-1)+2*(t-1)\leq 0\\(t-1)*(t^2+2)\leq 0\\t^2+2 0\ \ \ \ \Rightarrow\\t-1\leq 0\\t\leq 1\\t\geq 0\ \ \ \ \Rightarrow\\0\leq t\leq 1\ \ \ \ \ \Rightarrow\\\left \{ {{t\geq 0} \atop {t\leq 1}} \right. \ \ \ \ \ \left \{ {{\sqrt[6]{5x^2+24x+28}\geq 0 } \atop {\sqrt[6]{5x^2+24x+28}\leq 1 }} \right. \ \ \ \ \ \left \{ {{(\sqrt[6]{5x^2+24x+28})^6\geq 0^6 } \atop {(\sqrt[6]{5x^2+24x+28})^6\leq 1^6 }} \right.$

$\left \{ {5x^2+24x+28\geq 0} \atop {5x^2+24x+28\leq 1}} \right. \ \ \ \ \ \left \{ {5x^2+24x+28\geq 0} \atop {5x^2+24x+27\leq 0}} \right.\ \ \ \ \ \ \left \{ {{(x+2)*(5x+14)\geq 0} \atop {(x+3)*(5x+9)\leq 0}} \right. .$

1. (x+2)*(5x+14)≥0

-∞__+__-2,8__-__-2__+__+∞

x∈(-∞;-2,8]U[-2;+∞).

2. (x+3)*(5x+9)≤0

-∞__+__-3__-__-1,8__+__+∞

x∈[-3;-1,8].          ⇒

x∈[-3;-2,8]U[-2;-1,8].

Учитывая ОДЗ     ⇒

ответ: x∈[-3;-2,8]U[-2;-1,8].