Question

помогите Срочно решить по алгебре!!!!!

Answer 1

Ответ:

решение смотри на фотографии

Answer 2

Решение:

$2.\left \{ {{-4x+11 > 2x-7} \atop {8x-3\leq 6x+13}} \right. = > \left \{ {{-4x-2x > -7-11} \atop {8x-6x\leq 13+3}} \right.= > \left \{ {{-6x > -18} \atop {2x\leq 16}} \right.= > \left \{ {{x < 3} \atop {x\leq 8}} \right.= > x < 3\\\\x\in(-\infty;3)$

$3.f(x)=\sqrt{3x+7}+\frac{5}{\sqrt{8-4x}}\\\\\left \{ {{3x+7\geq 0} \atop {8-4x > 0}} \right.= > \left \{ {{3x\geq -7} \atop {4x < 8}} \right.= > \left \{ {{x\geq -7/3} \atop {x < 2}} \right.= > \left \{ {{x\geq -2\frac{1}{3}} \atop {x < 2}} \right.= > -2\frac{1}{3}\leq x < 2 \\\\x\in[-2\frac{1}{3};2)$