Question

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

Answer 1

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; +∞).