СРОЧНО РЕШИТЕ ХОТЯБЫ ОДНО

Question 1

Question 2

Решите задачу:

$1)\; \; | x^2-16x+36|\leq |36-x^2|\\\\\star \; \; |x|<a\; \;\Leftrightarrow \; \; -a<x<a\; \; \star \\\\-(36-x^2)\leq x^2-16x+36\leq 36-x^2\; \; \Leftrightarrow \; \; \; \left \{ {{x^2-36\leq x^2-16x+36} \atop {x^2-16x+36\leq 36-x^2}} \right. \\\\\left \{ {{16x-72\leq 0} \atop {2x^2-16x\leq 0}} \right. \; \left \{ {{x\leq 4,5} \atop {x(x-8)\leq 0}} \right. \; \left \{ {{x\leq 4,5} \atop {0\leq x\leq 8}} \right. \; \; \Rightarrow \; \; \underline {x\in [\, 0\, ;\, 4,5\, ]}\; -\; \; otvet.\\\\\star \; x(x-8)\leq 0\; ,\; \; znaki\, :\; \; \; +++[\, 0\, ]---[\, 8\, ]+++\; \; ,\; \; x\in [\, 0,8\, ]$

$2)\; \; \Big |\frac{x^2-5x+4}{x^2-4}\Big |\leq 1\; \; \Leftrightarrow \; \; -1\leq \frac{x^2-5x+4}{x^2-4}\leq 1\\\\\left \{ {{\frac{x^2-5x+4}{x^2-4}-1\leq 0} \atop {\frac{x^2-5x+4}{x^2-4}+1\geq 0} \right. \; \; \left \{ {{\frac{x^2-5x+4-x^2+4}{x^2-4}\leq 0} \atop {\frac{x^2-5x+4+x^2-4}{x^2-4}\geq 0}} \right. \; \left \{ {{\frac{-(5x-8)}{(x-2)(x+2)}\leq 0} \atop {\frac{2x^2-5x}{x^2-4}\geq 0}} \right. \; \left \{ {{\frac{5x-8}{(x-2)(x+2)}\geq 0} \atop {\frac{x(2x-5)}{(x-2)(x+2)}\geq 0}} \right.$

$a)\; \; \frac{5x-8}{(x-2)(x+2)}\geq 0:\; \; ---(-2)+++[\, 1,6\, ]---(2)+++\\\\x\in (-2;\, 1,6\, ]\cup (2,+\infty )\\\\b)\; \; \frac{x(2x-5)}{(x-2)(x+2)}\geq 0:\; \; +++(-2)---[\, 0\, ]+++(2)---[2,5\, ]+++\\\\x\in (-\infty ;-2)\cup[\, 0;2)\cup [\, 2,5;+\infty )\\\\Otvet:\; \; x\in [\, 0\, ;\, 1,6\, ]\cup [\, 2,5\, ;+\infty )\; .$