ОДЗ
{x≠0
{x+3>0⇒x>-3
{-x-1>0⇒x<-1<br>{x+6>0⇒x>-6
x∈(-3;-1)
перейдем к основанию 3
log(3)(x²)+log(3)(x+3)≥log(3)(-x-1)²+log(3)(x+6)
log(3)[x²(x+3)/(-x-1)²(x+6)]≥0
x²(x+3)/(-x-1)²(x+6)≥1
x²(x+3)/(x+1)²(x+6)-1≥0 (-x-1)²=(x+1)²
(x³+3x³-x³-6x³-2x³-12x-x-6)/(x+1)²(x+6)≥0
(-5x²-13x-6)/(x+1)²(x+6)≥0
(5x²+13x+6)/(x+1)²(x+6)≤0
5x²+13x+6=0
D=169-120=49
x1=(-13-7)/10=-2 U x2=(-13+7)/10=-0,6
x+1=0⇒x=-1
x+6=0⇒x=-6
\\\\\\\\\\\ /////////////////////////////////////
_ + + _ _ +
-------(-6)---------(-3)-------[-2]---------(-1)----------[-0,6]------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈[-2;-1)