Question

СРОЧНО!!!! Решите неравенство с помощью метода пробных точек!!!!!

Answer 1

|13 - 2x | ≥ |4x - 9|

(13 - 2x - (4x - 9))(13 - 2x + (4x - 9)) ≥ 0

(13 - 2x - 4x + 9)(13 - 2x + 4x - 9) ≥ 0

(22 - 6x)(2x + 4) ≥ 0

- 6(x - 22/6)* 2(x + 2) ≥ 0

(x - 11/3)(x + 2) ≤ 0

        +                      -                               +

_________[-2]________[3 2/3]____________

                     //////////////////////////

Ответ : x ∈ [ - 2 ; 3 2/3]

Answer 2

$1)\; \; |13-2x|\geq |4x-9|\\\\13-2x=0\; \; \to \; \; x=6,5\\\\4x-9=0\; \; \to \; \; x=2,25\\\\znaki\; (13-2x):\; \; +++(1,15)+++(6.5)---\\\\znaki\; (4x-9):\; \; \; \; ---(2,25)+++(6,5)+++\\\\a)\; \; x\leq 2,25:\; \; |13-2x|=13-2x\; ,\; |4x-9|=-(4x-9)=9-4x\; ,\\\\13-2x\geq 9-4x\; ,\; 2x\geq -4\; ,\; \; \underline {x\geq -2}\\\\\underline {x\in [-2\, ;\; 2,25\, ]}\\\\b)\; \; 2,25<x\leq 6,5:\; \; |13-2x|=13-2z\; ,\; |4x-9|=4x-9\; ,\\\\13-2x=4x-9\; ,\; \; 22\geq 6x\; ,\; \; \underline {x\leq 3\frac{2}{3}}$

$\underline {x\in (2,25\, ;\, 3\frac{2}{3}\, ]} \\\\c)\; \; x>6,5:\; \; |13-2x|=-(13-2z)=2x-13\; ,\; \; |4x-9|=4x-9\; ,\\\\2x-13\geq 4x-9\; ,\; \; -4\geq 2x\; ,\; \; x\leq -2\; \; ,\; \; -2\notin (6,5\, ;+\infty )\\\\Otvet:\; \; x\in [-2\, ;\, 3\frac{2}{3}\, ]\; .$