Question

Решите уравнения log4 ( x квадрат+x)=log4 (x квадрат+6)

Answer 1

{x²+x>0⇒x(x+1)>0  x=0  x=-1  x<-1 U x>0
{x²+6>0⇒x∈R
x∈(-∞;-1) U (0;∞)
x²+x=x²+6
x=6

Answer 2

Log₄(x²+x)=log₄(x²+6)
ОДЗ:
$\left \{ {{ x^{2} +x\ \textgreater \ 0} \atop { x^{2} +6\ \textgreater \ 0}} \right. \left \{ {{x*(x+1)\ \textgreater \ 0} \atop { x^{2} \ \textgreater \ -6}} \right.$
    +              -              +
--------(-1)----------(0)-------------->x
x∈(-∞;-1)∪(0;∞)
x²+x=x²+6
x=6