Question

Помогите, пожалуйста, решить неравенства:

$(x+2)(x-3)(x-4)\ \textless \ 0\\\\\frac{(x+5)(x-2)}{(x-1)^2} \geq 0\\\\\frac{2x+1}{x-3} \leq 1\\\\\frac{x}{x-4} + \frac{5}{x-1} + \frac{24}{x^2 -5x +4} \leq 0$

Answer 1

везде применяем метод интервалов

1.

(x+2)(x-3)(x-4) < 0

------------ (-2) +++++++++ (3) ------------- (4) ++++++++++

x∈(-∞ -2) U (3  4)

2

(x+5)/(x-2)/(x-1)^2 >=0

++++++++++ [-5] ------------ (1) ------------ [2] +++++++++++

x∈(-∞ -5] U [2  +∞)

3

(2x+1)/(x-3) <=1

(2x+1)/(x-3) - 1<=0

(2x+1 - x + 3)/(x-3)<=0

(x+4)/(x-3)<=0

+++++++ [-4] ------------- (3) +++++++

x∈[-4  3)

4

x/(x-4) + 5/(x-1) +   24/(x-1)(x-4) <=0

(x(x-1) + 5(x-4) + 24)/(x-1)(x-4) <=0

(x^2 - x + 5x - 20 + 24) /(x-1)(x-4) <=0

(x^2-4x+4)/(x-1)(x-4) <=0

(x-2)^2/(x-1)(x-4) <=0

++++++++ (1) -----------[2] ---------- (4) ++++++++

x∈(1 4)