Question

Помогите, пожалуйста ♡♡♡♡

2(x^2 + 2)> x (x + 5)

x-(x+4)(x+5)>-5

(6x-1)(6x+1)-(12x-5)(x+2)<7-3x

хоть что-то) )

Answer 1

1) 2(x² + 2) > x(x + 5)
2x² + 4 - x² - 5x > 0
x² - 5x + 4 > 0
(x - 4)(x - 1) > 0
    +                  -                   +
__________________________
           1                    4
x ∈ (- ∞ ; 1)∪(4 , + ∞)
2) x - (x + 4)(x + 5) > - 5
x - x² - 5x - 4x - 20 + 5 > 0
- x² - 8x - 15 > 0
x² + 8x + 15 < 0
(x + 5)(x + 3) < 0
            +                      -                          +
__________________________________
                    - 5                            - 3
x ∈ (- 5 ; - 3)
3) (6x - 1)(6x + 1) - (12x - 5)(x + 2) < 7 - 3x
36x² - 1 - 12x² - 24x + 5x + 10 - 7 + 3x < 0
24x² - 16x + 2 < 0
12x² - 8x + 1 < 0
12x² - 8x + 1 = 0
D = 64 - 4 * 12 * 1 = 64 - 48 =16
$X _{1}= \frac{8+ \sqrt{16} }{24} = \frac{8+4}{24}= \frac{12}{24}=0,5\\\\X _{2}= \frac{8- \sqrt{16} }{24}= \frac{8-4}{24}= \frac{4}{24} = \frac{1}{6} \\\\12(x-0,5)(x- \frac{1}{6}) \ \textless \ 0 \\\\(x-0,5)(x- \frac{1}{6})\ \textless \ 0$
      +                 -                      +
_____________________________
            1/6                   0,5
x ∈ (1/6 ; 0,5)

Answer 2

Решение задания приложено