Question

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Answer 1

Найдем ОДЗ:$\frac{7}{1-x} \ \textgreater \ 0$
1-x>0
x<1
$log_{ \frac{1}{3} } \frac{7}{1-x} \geq -1$
$-log_{3 } \frac{7}{1-x} \geq -1$
$log_{3 } \frac{7}{1-x} \leq 1$
$log_{3 } \frac{7}{1-x} \leq log_{3} 3$
$\frac{7}{1-x} \leq 3$
$\frac{7}{1-x}-3 \leq 0$
$\frac{7-3+3x}{1-x} \leq 0$
$\frac{4+3x}{1-x} \leq 0$
решаем методом интервалов
( - ∞;$1 \frac{1}{3}$] (1; + ∞)
в пересечении с ОДЗ
Ответ: ( - ∞; 1$\frac{1}{3}$]

Answer 2

Получится вот так :)

Answer 3

Simple Inequality

Sure, I can help you with that! To verify if you have solved a simple inequality correctly, I'll need you to provide the specific inequality that you want to check. Could you please provide the inequality that you would like me to verify?