Question

15 пожалуйста !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

ОДЗ
{x+3>0⇒x>-3
{log(1/5)(x+3)>0⇒x+3<1⇒x<-2
x∈(-3;-2)
log(1/5)(x+3)≥2
x+3≤0,04
x≤-2,96
x∈(-3;-2,96]

Answer 2

$ODZ: \\ \\ \left \{ {{x+3\ \textgreater \ 0} \atop { log_{ \frac{1}{5} }(x+3)\ \textgreater \ 0 }} \right. \\ \\ \left \{ {{x\ \textgreater \ -3} \atop {x\ \textless \ -2}} \right.$
x∈(-3;-2)

$log_{ \sqrt[6]{4} } ( log_{ \frac{1}{5} } (x+3)) \geq 3 \\ \\ log_{ \sqrt[3]{2} } ( log_{ \frac{1}{5} } (x+3)) \geq log_{ \sqrt[3]{2} } (\sqrt[3]{2})^3$

$\sqrt[3]{2} \ \textgreater \ 1$

знак сохраняем

$log_{ \frac{1}{5} } (x+3) \geq 2 \\ \\ log_{ \frac{1}{5} } (x+3) \geq log_{ \frac{1}{5} }( \frac{1}{5} )^2$

$\frac{1}{5} \ \textless \ 1$

знак меняем

$x+3 \leq ( \frac{1}{5} )^2 \\ \\ x+3 \leq 0.04 \\ \\ x \leq -2.96$

с учетом ОДЗ, получаем 

ОТВЕТ : x∈(-3;-2.96]