Решить неравенство (2x+1) (1-2x) (x-1) (2-3x) >0 (2-3x) (3x+2) (5+3x) (2x-3) >0 (3x-2) (5-x) (x+1) (2-x) <0

Найдем корни двучленов и решим неравенства:

a)

(2x + 1) (1 - 2x) (x - 1) (2 - 3x) > 0; (2x + 1) (2x - 1) (3x - 2) (x - 1) > 0; x1 = - 1/2; x2 = 1/2; x3 = 2/3; x4 = 1; x ∈ (-∞; - 1/2) ∪ (1/2; 2/3) ∪ (1; ∞).

b)

(2 - 3x) (3x + 2) (5 + 3x) (2x - 3) > 0; (3x + 5) (3x + 2) (3x - 2) (2x - 3) < 0; x1 = - 5/3; x2 = - 2/3; x3 = 2/3; x4 = 3/2; x ∈ (-5/3; - 2/3) ∪ (2/3; 3/2).

c)

(3x - 2) (5 - x) (x + 1) (2 - x) < 0; (x + 1) (3x - 2) (x - 2) (x - 5) < 0; x1 = - 1; x2 = 2/3; x3 = 2; x4 = 5; x ∈ (-1; 2/3) ∪ (2; 5).

Ответ:

a) (-∞; - 1/2) ∪ (1/2; 2/3) ∪ (1; ∞); b) (-5/3; - 2/3) ∪ (2/3; 3/2); c) (-1; 2/3) ∪ (2; 5).