Question

Решите неравенство методом интервалов

Answer 1

$(3x+7)log_{2x+5}(x^2+4x+5)\geq 0\\\left \{ {{2x+5>0} \atop {2x+5\neq 1}} \atop {x^2+4x+5>0}}\right.$

x∈(-5/2;+∞)\{-2}

$(3x+7)log_{2x+5}(x^2+4x+5)\geq 0<=>(3x+7)(2x+5-1)(x^2+4x+5-1)\geq 0\\(3x+7)(2x+4)(x^2+4x+4)\geq 0<=>(3x+7)(x+2)^3\geq 0\\--(+)--(-\frac{7}{3} )--(-)--(-2)--(+)--$

x∈(-∞;-7/3]∪[-2;+∞)

Наносим наше ограничение:

x∈(-5/2;-7/3]∪(-2;+∞)