Решите неравенство:

а) 2х(3х – 1) > 4х² + 5х + 9;

б) (5х + 7)(х – 2) < 21х² – 11х – 13.

Решение:

а) 2х(3х – 1) > 4х² + 5х + 9;

6x² – 2x > 4x² + 5x + 9

6x² – 2x – 4x² – 5x – 9 > 0

2x² – 7x – 9 > 0

D = (-7)² – 4 ⋅ 2 ⋅ (-9) = 49 + 72 = 121

x₁ = -(-7) – 11 / 2 ⋅ 2 = 7 – 11 / 4 = – 4/4 = -1

x₂ = -(-7) + 11 / 2 ⋅ 2 = 7 + 11 / 4 = 18/4 = 9/2 = 4,5

x (-∞; -1) U (4,5; +∞)

б) (5х + 7)(х – 2) < 21х² – 11х – 13.

5x² – 10x + 7x – 14 < 21x² – 11x – 13

5x² – 3x – 14 < 21x² – 11x – 13

5x² – 3x – 14 – 21x² + 11x + 13 < 0

– 16x² + 8x – 1 < 0

– 16x² + 8x – 1 = 0

16x² – 8x + 1 = 0

(4x – 1)² = 0

4x – 1 = 0

4x = 1

x = 1: 4

x = 0,25

x (-∞; 0,25) U (0,25; + ∞)

Источник : Учебник по Алгебре 9 класса авторов Ю.Н. Макарычев, Н.Г. Миндюк, К.И. Нешков, С.Б. Суворова, 2014г.