СРОЧНО РЕШИТЕ ХОТЯБЫ ОДНО. Created by Polina2100Kaddik. algebra-ru

СРОЧНО РЕШИТЕ ХОТЯБЫ ОДНО...

Polina2100Kaddik @Polina2100Kaddik

August 2022 1 6 Report

СРОЧНО РЕШИТЕ ХОТЯБЫ ОДНО

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NNNLLL54

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$1)\; \; | x^2-16x+36|\leq |36-x^2|\\\\\star \; \; |x|$

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

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

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