ОДЗ
{x≠0
{x+14>0⇒x>-14
{x+4>0⇒x>-4
{x+2>0⇒x>-2
x∈(-2;0) U (0;∞)
Перейдем к основанию 2
log(2)(x²)+log(2)(x+14)≤log(2)(x+4)+log(2)(x+2)²
log(2)(x²)+log(2)(x+14)-log(2)(x+4)-log(2)(x+2)²≤0
log(2)[x²(x+14)/[(x+4)(x+2)²]]≤0
x²(x+14)/[(x+4)(x+2)²]≤1
x²(x+14)/[(x+4)(x+2)²]-1≤0
(x³+14x²-x³-4x²-4x-4x²-16x-16)/[(x+4)(x+2)]≤0
(6x²-20x-16)/[(x+4)(x+2)²]≤0
6x²-20x-16=0
3x²-10x-8=0
D=100+96=196
x1=(10-14)/6=-2/3 U x2=(10+14)/6=4
x+2=0⇒x=-2
x+4=0⇒x=-4
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x∈[-2/3;0) U (0;4]