Question

Разрешите неравенство |x+2|+|x-1|-|x-4|> 3

Answer 1

$|x+2|+|x-1|-|x-4|3\\\\x+2=0\; \to \; x_1=-2\; \; ;\; \; x-1=0\; \to \; x_2=1\; \; ;\; \; x-4=0\; \to \; x_3=4\\\\a)\; x\leq -2:\; |x+2|=-(x+2)\; ;\; |x-1|=-(x-1)\; ;\; |x-4|=-(x-4);\\\\-x-2-x+1+x-43\; ;\; -x8\; \to \; \underline {x4\; ;\\\\\left \{ {{-24}} \right. \; \; \Rightarrow \; \; \underline {x\in \varnothing }\\\\c)\; 13\; ;\; 3x6\; ;\; x2\; ;$

$\left \{ {{12}} \right. \; \; \to \; \; \underline {24:\; |x+2|=x+2\; ;\; |x-1|=x-1\; ;\; |x-4|=x-4;\\\\x+2+x-1-x+43\; ;\; x-2\; ;\\\\\left \{ {{x4} \atop {x-2}} \right. \; \; \to \; \; \underline {x4}\\\\Otvet:\; \; x\in (-\infty ,-8)\cup (2,4\, ]\cup (4,+\infty )\; \Rightarrow \; x\in (-\infty ,-8)\cup (2,+\infty )\; .$