Question

разрешите неравенство |x+2|+|x-1|-|x-4|>3 даю 34бала

Answer 1

$|x+2|+|x-1|-|x-4|>3\\\\x+2=0\; \to \; x_1=-2\; \; ;\; \; x-1=0\; \to \; x_2=1\; \; ;\; \; x-4=0\; \to \; x_3=4\\\\a)\; x\leq -2:\; |x+2|=-(x+2)\; ;\; |x-1|=-(x-1)\; ;\; |x-4|=-(x-4);\\\\-x-2-x+1+x-4>3\; ;\; -x>8\; \to \; \underline {x<-8}\\\\b)\; -2<x\leq 1:\; |x+2|=x+2\; ;\; |x-1|=-(x-1)\; ;\; |x-4|=-(x-4);\\\\x+2-x+1+x-4>3\; ;\; x>4\; ;\\\\\left \{ {{-2<x\leq 1} \atop {x>4}} \right. \; \; \Rightarrow \; \; \underline {x\in \varnothing }\\\\c)\; 1<x\leq 4:\; |x+2|=x+2\; ;\; |x-1|=x-1\; ;\; |x-4|=-(x-4);\\\\x+2+x-1+x-4>3\; ;\; 3x>6\; ;\; x>2\; ;$

$\left \{ {{1<x\leq 4} \atop {x>2}} \right. \; \; \to \; \; \underline {2<x\leq 4}\\\\d)\; x>4:\; |x+2|=x+2\; ;\; |x-1|=x-1\; ;\; |x-4|=x-4;\\\\x+2+x-1-x+4>3\; ;\; x>-2\; ;\\\\\left \{ {{x>4} \atop {x>-2}} \right. \; \; \to \; \; \underline {x>4}\\\\Otvet:\; \; x\in (-\infty ,-8)\cup (2,4\, ]\cup (4,+\infty )\; \Rightarrow \; x\in (-\infty ,-8)\cup (2,+\infty )\; .$