Question

Срочно даю 35 балов
a) X² - 6x<0;
b) 8x + x² >= 0;
c) x² =< 4;
d) x² > 6.

Answer 1

$a)x^2-6x<0\\x\cdot(x-6)<0\\\left \{ {{x<0} \atop {x-6>0}} \right.\left \{ {{x>0} \atop {x-6<0}} \right.\\\left \{ {{x<0} \atop {x>6}} \right.\left \{ {{x>0} \atop {x<6}} \right. \\x\in\varnothing\\x\in(0,6)\\Otvet: x\in(0,6)$
$b)8x+x^2\geq0\\x\cdot(8+x)\geq0\\\left \{ {{x\geq0} \atop {8+x\geq0}} \right. \left \{ {{x\leq0} \atop {8+x\leq0}} \right. \\\left \{ {{x\geq 0} \atop {x\geq -8}} \right. \left \{ {{x\leq0} \atop {x\leq -8}} \right. \\ x\in[0;+\infty)\\x\in(-\infty;-8]\\x\in(-\infty;-8]\cup[0;+\infty)$
$d)x^2>6\\|x|>\sqrt{6}\\x>\sqrt{6},x\geq0\\-x>\sqrt{6},x<0\\x\in(\sqrt{6};+\infty)\\x<-\sqrt{6};x<0)\\x\in(\sqrt{6};+\infty)\\x\in(-\infty;-\sqrt{6})\\x\in(-\infty;-\sqrt{6})\cup(\sqrt{6};+\infty)$