Решите неравенства: 5х2+3х-8 больше 0 (2х2-3х+1)(х-3) больше =0 х2-2х-15 больше =0 2х+3/х+2 меньше 1 (5х+4)(3х-2)/х+3 меньше =(3х-2)(х+2)/1-х

5x² + 3x - 8 > 0 5x² + 3x - 8 = 0 D = 9 + 8·4·5 = 169 = 13² [latex]x_1 = \dfrac{-3 + 13}{10} = 1 \\ \\ x_2 = \dfrac{-3 - 13}{10} = -1,6 [/latex] 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 [latex]x_1 = \dfrac{3 + 1}{4} =1 \\ \\ x_2 = \dfrac{3 - 1}{4} = 0,5[/latex] 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; +∞) [latex] \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) [/latex] [latex] \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 [/latex] [latex] \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[/latex] [latex]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 [/latex] [latex]\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 [/latex] 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; +∞).