Question

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Answer 1

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

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