Ответ:
x∈(-∞,0)∪[1/3,+∞), x∈(0,2/3)
x∈[1/3,2/3)
Объяснение: 2㏒1/9(2-3x/x)≥-1
x∈(0,2/3)
㏒1/9(2-3x/x)≥-1/2
2-3x/x≤(1/9)^-1/2
2-3x/x≤9^1/2
2-3x/x≤(3²)^1/2
2-3x/x≤3
(2-3x/x)-3≤0
2-3x-3x/x≤0
2-6x/x≤0
2(1-3x)/x≤0
0} \right. \\\left \{ {{2(1-3x)\leq 0 } \atop {x<0}} \right. \left \{ {{x\geq \frac{1}{3} } \atop {x>0}} \right. \left \{ {{x\leq\frac{1}{3} } \atop {x<0}} \right." alt="\left \{ {{2(1-3x)\leq 0 } \atop x>0} \right. \\\left \{ {{2(1-3x)\leq 0 } \atop {x<0}} \right. \left \{ {{x\geq \frac{1}{3} } \atop {x>0}} \right. \left \{ {{x\leq\frac{1}{3} } \atop {x<0}} \right." align="absmiddle" class="latex-formula">