Question

30 БАЛЛОВ.Помогите решить неравенство (9 класс):
$| {x }^{2} - 12x + 25 | < 3x - 11$
Ответ расписать​

Answer 1

|x² - 12x + 25| < 3x - 11
[-x² + 12x - 25 < 3x - 11
[-x² + 12x - 25 < -3x + 11,

[-x² + 9x - 14 < 0
[-x² + 15x - 36 < 0,

[x² - 9x + 14 > 0
[x² - 15x + 36 > 0,

[(x - 2)(x - 7) > 0
[(x - 12)(x - 3) > 0,

[x ∈ (-∞; 2) ∪ (7; +∞)
[x ∈ (-∞; 3) ∪ (12; +∞)

3x - 11 > 0
x > 11/3

OTVET: x ∈ (7; +∞)

Answer 2

$\boxed {\; \; |x|<a\; \; \Leftrightarrow \; \; \; -a<x<a\; }$

$|x^2-12x+25|<3x-11\; \; \Rightarrow \; \; \; -(3x-11)<x^2-12x+25<3x-11\\\\\\\left \{ {{x^2-12x+25\, <\, 3x-11} \atop {x^2-12x+25\, >\, 11-3x}} \right.\; \; \left \{ {{x^2-15x+36\, <0} \atop {x^2-9x+14\, >0}} \right. \; \; \left \{ {{(x-3)(x-12)\, <0} \atop {(x-2)(x-7)\, >0}} \right.\; \; \left \{ {{+++(3)---(12)+++} \atop {+++(2)---(7)+++}} \right. \\\\\\\left \{ {{x\in (3\, ;\, 12)\qquad \qquad } \atop {x\in (-\infty ,2)\cup (7,+\infty )\; }} \right.\; \; \; \Rightarrow \; \; \; x\in (7\, ;\, 12)$