Question

Помогите решить! Срочно!

Answer 1

bb$left { {{4-2x>0} atop {x+2>0}} right. left { {{x<2} atop {x>-2}} right.$

$log_{0,3}frac{4-2x}{x+2} geq 0,; ; 0=log_{0,3}1\\frac{4-2x}{x+2}-1 leq 0\\frac{2-3x}{x+2} leq 0,; ; + + + (-2)- - - -[frac{2}{3}+ + + +\\xin (-2,frac{2}{3}]\\Otvet:xin (-2,frac{2}{3}]$