Question

Решить систему неравенств.

Answer 1

$ODZ:\; x\ne 1,2,3,4\\\\ |2x-5| \geq 3\; \; \to \; \; 2x-5 \geq 3\; \; ili\; \; 2x-5 \leq -3\\\\ \left [ {{2x\geq 8} \atop {2x\leq 2}} \right. \; \left \{ {{x\geq 4} \atop {x\leq 1}} \right.\; \; \to \; \; x\in (-\infty,1\, ]U[\, 4,+\infty) \\\\Tak\; kak\; vse\; vurazenija\; pod\; modylyami\; pri\; x\in (-\infty,1\, ]\; otricatelnu,\\\\a\; pri\; x\in [\, 4,+\infty)\; oni\; polozitelnu, to\; imeem \; 2 \; slychaja:\\\\1)x\in ( 4,+\infty),\\\\|x-1|=x-1,\; |x-2|=x-2,\; |x-3|=x-3,\; |x-4|=x-4$

$\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}\geq 0,3\\\\Formyla:\; \frac{1}{n(n-1)}=\frac{-1}{n}+\frac{1}{n-1}\\\\-\frac{1}{x-1}+\frac{1}{x-2}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}\geq 0,3\\\\\frac{1}{x-4}-\frac{1}{x-1}\geq 0,3\\\\\frac{x-1-x+4-0,3(x^2-5x+4)}{(x-1)(x-4)}\geq 0\\\\\frac{3-0,3(x^2-5x+4)}{(x-1)(x-4)}\geq 0\\\\\frac{-0,3(-10+ x^2-5x+4)}{(x-1)(x-4)}\geq 0\\\\\frac{x^2-5x-6}{(x-1)(x-4)}\leq 0\\\\x\in [\, -1,1)U(4,6\, ]\\\\x\in (4,6\, ]$

$2)\; x\in (-\infty ,1)\\\\|x-1|=-(x-1),\; |x-2|=-(x-2),\; |x-3|=-(x-3),\; |x-4|=-(x-4)\\\\\frac{1}{(x-1)(x-2}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}\geq 0,3\\\\x\in [\, -1,1)U(4,6\, ]\\\\x\in [-1,1)\\\\Otvet:\; x\in [\, -1,1)U( 4,6\, ]$