Question

htibnt ehfdytybtlog0.2(x^2+4x)= -1

Answer 1

ОДЗ
х²+4х>0
x(x+4)>0
x∈(-∞; -4)∨(0; +∞)

$log_{0.2}( x^{2}+4x)=-1 \ log_{ frac{1}{5} }( x^{2}+4x)=-1 \ x^{2} +4x=( frac{1}{5})^{-1} \ x^{2} +4x =5 \ x^{2} +4x-5=0 \ D=4^2+4*5=36 \ \ x_1= frac{-4+6}{2}=1 \ x_2= frac{-4-6}{2}=-5$