Задать вопрос Регистрация
Помочь другим

  • Допоможіть розвязати нерівності*)) що зможите))
    1) log 3 (2x+1) < log 3 ( x-1)
    2)log 1/3 (2-x ) > 0
    3) log 31 ( 31x+2) <1
    4) log 1/11 ( 2x-1) + log1/11 x>0

Ответы 1

  • 1) log_3(2x+1)\ \textless \ log_3(x-1)\\3\ \textgreater \ 1 =\ \textgreater \ 2x+1\ \textless \ x-1\\x\ \textless \ -2\\D: \left \{ {{2x+1\ \textgreater \ 0} \atop {x-1\ \textgreater \ 0}} ight. ,\left \{ {{x\ \textgreater \ -\frac{1}{2}} \atop {x\ \textgreater \ 1}} ight. , x\ \textgreater \ 1\\ \\ \left \{ {{x\ \textless \ -2} \atop {x\ \textgreater \ 1}} ight. нет решений2) log_{\frac{1}{3}}(2-x)\ \textgreater \ 0\\log_{\frac{1}{3}}(2-x)\ \textgreater \ log_{\frac{1}{3}}1\\\frac{1}{3}\ \textless \ 1 =\ \textgreater \ 2-x\ \textless \ 1\\x\ \textgreater \ 1\\ \\D: 2-x\ \textgreater \ 0, x\ \textless \ 2\\ \\ \left \{ {{x\ \textgreater \ 1} \atop {x\ \textless \ 2}} ight. 1\ \textless \ x\ \textless \ 23) log_{31}(31x+2)\ \textless \ 1\\log_{31}(31x+2)\ \textless \ log_{31}31\\31\ \textgreater \ 1 =\ \textgreater \ 31x+2\ \textgreater \ 31\\31x\ \textgreater \ 29\\x\ \textgreater \ \frac{29}{31}\\ \\ D: 31x+2\ \textgreater \ 0, 31x\ \textgreater \ -2, x\ \textgreater \ -\frac{2}{31}\\ \\ \left \{ {{x\ \textgreater \ \frac{29}{31}} \atop {x\ \textgreater \ -\frac{2}{31}}} ight. x\ \textgreater \ \frac{29}{31}4) log_{\frac{1}{11}}(2x-1)+log_{\frac{1}{11}}x\ \textgreater \ 0\\log_{\frac{1}{11}}(2x-1)\ \textgreater \ -log_{\frac{1}{11}}x\\log_{\frac{1}{11}}(2x-1)\ \textgreater \ log_{\frac{1}{11}}x^{-1}\\log_{\frac{1}{11}}(2x-1)\ \textgreater \ log_{\frac{1}{11}}\frac{1}{x}\\\frac{1}{11}\ \textless \ 1=\ \textgreater \ 2x-1\ \textless \ \frac{1}{x}\\\frac{2x^2-x-1}{x}\ \textless \ 0\\\frac{2(x-1)(x+\frac{1}{2})}{x}\ \textless \ 0\\x\ \textless \ -\frac{1}{2}, x\ \textgreater \ 1D: \left \{ {{2x-1\ \textgreater \ 0} \atop {x\ \textgreater \ 0}} ight. \\ \left \{ {{x\ \textgreater \ \frac{1}{2}} \atop {x\ \textgreater \ 0}} ight. ,x\ \textgreater \ \frac{1}{2}\\ \\ \left \{ {{x\ \textless \ -\frac{1}{2}, x\ \textgreater \ 1} \atop {x\ \textgreater \ \frac{1}{2}}} ight. , x\ \textgreater \ 1

    • Автор:

      dunnp6fk
    • 6 лет назад
    • 0

  • Добавить свой ответ

Еще вопросы

Топ пользователей

Войти через Google

 или

Забыли пароль?

У меня нет аккаунта, я хочу Зарегистрироваться

How much to ban the user?
1 hour 1 day 100 years