Question

Реши два неравенства
98б

Answer 1

$1)\; \; |x^2-x+1|\geq |x^2-3x+4|\; \; \to \; \; |x^2-3x+4|\leq |x^2-x+1|\\\\\star \; \; |A|<B\; \; \Rightarrow \; \; -B<A<B\; \; \star \\\\-(x^2-x+1)\leq x^2-3x+4\leq x^2-x+1\; \; \Rightarrow \left \{ {{x^2-3x+4\leq x^2-x+1} \atop {x^2-3x+4\geq -x^2+x-1}} \right. \\\\a)\; \; x^2-3x+4\leq x^2-x+1\; \; ,\; \; 3\leq 2x\; ,\; \; x\geq \frac{3}{2}\\\\b)\; \; x^2-3x+4\geq -x^2+x-1\; ,\; \; 2x^2-4x+5\geq 0\; ,\\\\D/4=4-10=-6<0\; \; \Rightarrow \; \; 2x^2-4x+5>0\; \; pri\; \; x\in (-\infty ,+\infty )\\\\Otvet:\; x\in [\, \frac{3}{2}\, ,\, +\infty )$

$2)\; \; |x^2+x-6|<x\\\\-x<x^2+x-6<x\; \; \Rightarrow \; \; \left \{ {{x^2+x-6<x} \atop {x^2+x-6>-x}} \right. \; \left \{ {{x^2-6<0} \atop {x^2+2x-6>0}} \right. \\\\a)\; \; x^2-6<0\; ,\; \; (x-\sqrt6)(x+\sqrt6)<0\; ,\; \sqrt6\approx 2,4\\\\znaki:\; \; +++(-\sqrt6)---(\sqrt6)+++\quad x\in (-\sqrt6,+\sqrt6)\\\\b)\; \; x^2+2x-6>0\; ,\; \; D/4=1+6=7\; ,\; x_{1,2}=-1\pm \sqrt7\\\\(x+1-\sqrt7)(x+1+\sqrt7)>0\\\\znaki:\; \; \; +++(-1-\sqrt7)---(-1+\sqrt7)+++\\\\x\in (-\infty ,-1-\sqrt7)\cup (-1+\sqrt7,+\infty )$

$-1-\sqrt7\approx -3,6\; \; ;\; \; -1+\sqrt7\approx 1,6\\\\c)\; \; \left \{ {{x\in (-\sqrt6,\sqrt6)} \atop {x\in (-\infty ,-1-\sqrt7)\cup (-1+\sqrt7,+\infty )}} \right. \; \; \Rightarrow \; \; x\in (-1+\sqrt7\, ;\, \sqrt6)$