Question

1) (x^2-3x+24)/(x^2-3x +3)<4 2)(x^2-4x)/8+(x-3)/5>=(1-x)/6 3) (2x^2-x-3)(2x^2-9x-9)>=0 4) (x^2-6x+9)/(5-4x-x^2)>=0 метод интервалов

Answer 1

В чём суть метода интервалов? интервалы - это числовые промежутки. Откуда их взять? надо найти "нули" всех компонентов неравенства. потом на каждом интервале надо определить знак.

1) (x^2-3x+24)/(x^2-3x +3)<4

(x^2-3x+24)/(x^2-3x +3) - 4 < 0

(x^2-3x+24 - 4* (x^2-3x +3) ) / (x² -3x +3) < 0

(х² -3x +24 -4x² + 12х -12 )/(х² -3х +3) < 0

(-3x² +9x + 12)/(x² -3x +3) < 0

-3x² +9x + 12 = 0 или x² -3x +3= 0

х² -3х - 4 = 0 ∅

корни 4 и -1

-∞ -1 4 +∞

- + - это знаки -3x² +9x + 12

Ответ: х ∈ (-1; 4)

2)(x^2-4x)/8+(x-3)/5 ≥ (1-x)/6 | * 120

15( x² -4x) + 24*( x - 3) ≥ 20(1 - x)

15x² - 60x +24x -72 -20 +20x ≥ 0

15x² - 16x -92 ≥ 0

x = (8 +-√(64 + 1380) )/ 15 = (8 +-38)/15

х₁ = 46/15; х₂ = -2

-∞ -2 46/15 +∞

+ - + это знаки 15x² - 16x -92

Ответ: х∈ (-∞; -2] ∪ [ 46/15; +∞)

3) (2x^2-x-3)(2x^2-9x-9) ≥ 0

(2x^2-x-3) = 0 или (2x^2-9x-9) = 0

корни 1,5 и -1 (9 + √153)/4 и (9 -√153)/4

-∞ -1 (9 -√153)/4 1,5 (9 +√153)/4 + ∞

+ - - + + знаки (2x^2-x-3)

+ + - - + знаки 2x^2-9x-9

IIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIi IIIIIIIIIIIIII решение

Ответ: (-∞; -1] ∪[(9 -√153)/4 ; 1,5] ∪[(9 +√153)/4 ; + ∞)

4) (x^2-6x+9)/(5-4x-x^2) ≥ 0

(x^2-6x+9) = 0 или (5-4x-x^2)

корни 3 корни -5 и 1

-∞ -5 1 3 +∞

+ + + + знаки x^2-6x+9

- + - - знаки 5-4x-x^2

IIIIIIIIIIIIIIIIIIII решение

Ответ: х∈ [-5;1 ]

Answer 2

I see you've got some inequalities to solve! Let's tackle them one by one:

1. $\frac{x^2-3x+24}{x^2-3x+3} < 4$

First, factor the denominators: $\frac{(x-8)(x+3)}{(x-1)(x-3)} < 4$

Now, find the critical points where the expression is undefined or equal to zero: $x = 1, 3$

Create intervals using these points and test points within each interval to determine the sign of the expression. The solution will be where the expression is less than zero.

2. $\frac{x^2-4x}{8} + \frac{x-3}{5} \geq \frac{1-x}{6}$

Combine the fractions on the left side: $\frac{5(x^2-4x) + 8(x-3)}{40} \geq \frac{1-x}{6}$

Simplify and solve for $x$. Remember to consider the values that make the denominators zero.

3. $(2x^2-x-3)(2x^2-9x-9) \geq 0$

Factor the quadratic and find the critical points. Use test points in the intervals created by these points to determine the sign of the expression.

4. $\frac{x^2-6x+9}{5-4x-x^2} \geq 0$

Factor the denominator and simplify the expression. Find the critical points and test points to determine the intervals where the expression is greater than or equal to zero.

It's a bit of work, but you've got this!