Решить неравенства (во вложении):

Question 1

Question 2

Рациональное решение

Question 3

4) $x^2-4|x|\ \textless \ 12$
Рассмотрим два случая:
$\left \{ {{x \geq 0} \atop {x^2-4x-12\ \textless \ 0}} \right. \\ \left \{ {{x \geq 0} \atop {(x-6)(x+2)\ \textless \ 0}} \right. \\ \left \{ {{x \geq 0} \atop {x \in (-2; 6)}} \right. \\ x\in[0;6)$ и $\left \{ {{x\ \textless \ 0} \atop {x^2+4x-12\ \textless \ 0}} \right. \\ \left \{ {{x\ \textless \ 0} \atop {(x-2)(x+6)\ \textless \ 0}} \right. \\ \left \{ {{x\ \textless \ 0} \atop {x\in(-6; 2)}} \right. \\ x\in(-6; 0)$
Отсюда x∈(-6; 6)

5) $x^2-5x+9\ \textgreater \ |x-6|$
Опять рассмотрим два случая:
$\left \{ {{x-6 \geq 0} \atop {x^2-5x+9\ \textgreater \ x-6}} \right. \\ \left \{ {{x \geq 6} \atop {x^2-6x+15\ \textgreater \ 0}} \right. \\ \left \{ {{x \geq 6} \atop {x\in(-\infty; +\infty)}} \right. \\ x\in[6; +\infty)$ и $\left \{ {{x-6\ \textless \ 0} \atop {x^2-5x+9 \ \textgreater \ 6 - x}} \right. \\ \left \{ {{x\ \textless \ 6} \atop {x^2-4x+3\ \textgreater \ 0}} \right. \\ \left \{ {{x\ \textless \ 6} \atop {(x-3)(x-1)\ \textgreater \ 0}} \right. \\ \left \{ {{x\ \textless \ 6} \atop {x\in(-\infty; 1)\cup(3;+\infty)}} \right. \\ x\in(-\infty;1)\cup(3;6)$
Отсюда x∈(-∞; 1)∪(3; +∞)

6. $x^2+2|x-1|+7 \leq 4|x-2|$
Здесь уже рассмотрим 3 случая - x относительно чисел 1 и 2.
$\left \{ {{x \geq 2} \atop {x^2+2(x-1)+7 \leq 4(x-2)}} \right. \\ \left \{ {{x \geq 2} \atop {x^2+2x-2+7-4x+8 \leq 0}} \right. \\ \left \{ {{x \geq 2} \atop {x^2-2x+13 \leq 0}} \right. \\ \left \{ {{x \geq 2} \atop {x\in\varnothing}} \right. \\ x\in\varnothing$ $\left \{ {{1 \leq x \ \textless \ 2} \atop {x^2+2(x-1)+7 \leq -4(x-2)}} \right. \\ \left \{ {{1 \leq x \ \textless \ 2} \atop {x^2+2x-2+7+4x-8 \leq 0}} \right. \\ \left \{ {{1 \leq x \ \textless \ 2} \atop {x^2+6x-3 \leq 0}} \right. \\ \left \{ {{1 \leq x \ \textless \ 2} \atop {(x+3-2 \sqrt{3})(x+3+2 \sqrt{3}) \leq 0}} \right. \\ \left \{ {{1 \leq x \ \textless \ 2} \atop {x\in[-3-2 \sqrt{3}; -3+2 \sqrt{3}] }} \right. \\ x\in\varnothing$ $\left \{ {{x\ \textless \ 1} \atop {x^2-2(x-1)+7 \leq -4(x-2)}} \right. \\ \left \{ {{x\ \textless \ 1} \atop {x^2-2x+2+7+4x-8 \leq 0}} \right. \\ \left \{ {{x\ \textless \ 1} \atop {x^2+2x+1 \leq 0}} \right. \\ \left \{ {{x\ \textless \ 1} \atop {x=-1}} \right. \\ x=-1$
Отсюда x=-1