Question

 ˃ 0

Answer 1

(64x-x³)/5x>0
x(64-x²)/5x>0. x≠0
(64-x²)/5>0. 5>0, ⇒64-x²>0
8²-x²>0
(8-x)*(8+x)>0
метод интервалов:
1. (8-х)*(8+х)=0
8-х=0 или 8+х=0
х=8, х=-8

2.
           -             +                -
---------------|--------------|------------------->x
                 -8              8
x∈(-8;8)

Answer 2

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