Найдите множество решений неравенства:

а) Зх² + 40х + 10 < -х² + 11х + 3;

б) 9х² – х + 9 >= 3х² + 18х – 6;

в) 2х² + 8х – 111 < (3х – 5)(2х + 6);

г) (5х + 1)(3х – 1) > (4х – 1)(х + 2).

Решение:

а) Зх² + 40х + 10 < -х² + 11х + 3;

3x² + 40x + 10 + x² – 11x – 3 < 0

4x² + 29x + 7 < 0

D = 29² – 4 ⋅ 4 ⋅ 7 = 841 – 112 = 729 > 0

√D = √729 = √27² = 27

x₁ = -29 – 27/2 ⋅ 4 = -(29 + 27) / 8 = – 56/8 = -7

x₂ = -29 + 27/2 ⋅ 4 = -(29 – 27) / 8 = – 2/8 = – 0.25

x (-7; -0,25)

б) 9х² – х + 9 >= 3х² + 18х – 6;

9x² – x + 9 – 3x² – 18x + 6 >= 0

6x² – 19x + 15 >= 0

D = (-19)² – 4 ⋅ 6 ⋅ 15 = 361 – 360 = 1

√D = √1 = 1

x₁ = -(-19) – 1/2 ⋅ 6 = 19 – 1 / 12 = 18/12 = 3/2 = 1,5

x₂ = -(-19) + 1/2 ⋅ 6 = 19 + 1 / 12 = 20/12 = 5/3 = 1 2/3

x (-∞:1,5) U (1 2/3; +∞)

в) 2х² + 8х – 111 < (3х – 5)(2х + 6);

2x² + 8x – 111 < 6x² + 18x – 10x – 30

2x² + 8x – 111 < 6x² + 8x – 30

2x² + 8x – 111 – 6x² – 8x + 30 < 0

-4x² – 81 < 0

-4x² – 81 = 0

-4x² = 81

x² = 81 : (-4)

x² = – 20,25 – корней нет

x (-∞; +∞)

г) (5х + 1)(3х – 1) > (4х – 1)(х + 2)

15x² – 5x + 3x – 1 > 4x² + 8x – x – 2

15x² – 2x – 1 > 4x² + 7x – 2

15x² – 2x – 1 – 4x² – 7x + 2 > 0

11x² – 9x + 1 > 0

D = (-9)² – 4 ⋅ 11 ⋅ 1 = 81 – 44 = 37 > 0

√D = √37

x₁ = -(-9) – √37 / 2 ⋅ 11 = 9 – √37 / 22

x₂ = -(-9) + √37 / 2 ⋅ 11 = 9 + √37 / 22

x (-∞; 9 – √37/22) U (9 + √37/22; +∞)

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