Опубликовано 5 лет назад по предмету Алгебра от alina281998

решите неравенства: 5х2+3х-8>0

(2х2-3х+1)(х-3)>=0

х2-2х-15>=0

2х+3/х+2<1

(5х+4)(3х-2)/х+3<=(3х-2)(х+2)/1-х

Ответ

Ответ дан Dимасuk

5x² + 3x - 8 > 0
5x² + 3x - 8 = 0
D = 9 + 8·4·5 = 169 = 13²

$x_1 = dfrac{-3 + 13}{10} = 1 \ \ x_2 = dfrac{-3 - 13}{10} = -1,6$
5(x - 1)(x + 1,6) > 0
(x - 1)(x + 1,6) > 0
x ∈ (-∞; -1,6) U (1; +∞)

(2x² - 3x + 1)(x - 3) ≥ 0
2x² - 3x + 1 = 0
D = 9 - 2·4 = 1

$x_1 = dfrac{3 + 1}{4} =1 \ \ x_2 = dfrac{3 - 1}{4} = 0,5$

2(x - 1)(x - 0,5)(x - 3) ≥ 0
(x - 1)(x - 0,5)(x - 3) ≥ 0
- 0,5 + 1 - 3 +
-------------• ---------------• --------------------------• -----------> x
x ∈ [0,5; 1] U [3; +∞)

x² - 2x - 15 ≥ 0
x² - 2x + 1 - 4² ≥ 0
(x - 1)² - 4² ≥ 0
(x - 1 - 4)(x - 1 + 4) ≥ 0
(x - 5)(x + 3) ≥ 0
x ∈ (-∞; -3] U [5; +∞)

$dfrac{2x + 3}{x + 2} textless 1 \ \ dfrac{2x + 3}{x + 2} - 1 textless 0 \ \ dfrac{2x + 3}{x + 2} - dfrac{x+ 2 }{x + 2} textless 0 \ \ dfrac{2x + 3 - x - 2 }{x + 2} textless 0 \ \ dfrac{x + 1 }{x + 2} textless 0 \ \ x in (-2; -1)$

$dfrac{(5x + 4)(3x - 2)}{x + 3} leq dfrac{(3x - 2)(x + 2)}{x - 1 } \ \ dfrac{(5x + 4)(3x - 2)}{x + 3} - dfrac{(3x - 2)(x + 2)}{x - 1 } leq 0 \ \ \ dfrac{(5x + 4)(3x - 2)(x - 1)}{(x + 3)(x - 1)} - dfrac{(3x - 2)(x + 2)(x + 3)}{(x - 1 )(x + 3)} leq 0 \ \ \ dfrac{(5x + 4)(3x - 2)(x - 1) - (3x - 2)(x + 2)(x + 3)}{(x - 1)(x + 3)} leq 0$

$dfrac{(3x - 2)((5x + 4)(x - 1) - (x + 2)(x + 3))}{(x- 1)(x + 3)} leq 0 \ \ dfrac{(3x - 2)(5x^2- 5x + 4x - 4 - (x^2 + 3x + 2x + 6) }{(x - 1)(x + 3)} leq 0 \ \ dfrac{(3x - 2)(5x^2 - x - 4 - x^2 - 5x - 6)}{(x - 1)(x + 3)} leq 0 \ \ dfrac{(3x - 2)(4x^2 - 6x - 10)}{(x - 1)(x + 3)} leq 0 \ \ dfrac{(3x - 2)(2x^2 - 3x - 5)}{(x - 1)(x + 3)} leq 0$

$2x^2 - 3x - 5 = 0 \ \ D = 9 + 4 cdot 5 cdot 2 = 49 = 7^2 \ \ x_1 = dfrac{3 + 7}{4} = 2.5 \ \ x_2 = dfrac{3 - 7}{4} = -1$

$dfrac{(3x - 2)(x - 2,5)(x + 1))}{(x - 1)(x + 3)} leq 0 \ \ \ dfrac{(x - dfrac{2}{3})(x - 2,5)(x + 1) }{(x - 1)(x + 3)} leq 0$

Нули числителя: x = -1; 2/3; 2,5.
Нули знаменателя: x = -3; 1
- -3 + -1 - 2/3 + 1 - 2,5 +
----°-------------• -------------• ----------------°-------------------• ------------> x
Ответ: x ∈ (-3; -1] U [2/3; 1) U [2,5; +∞).