Іть 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

Ответ:

nikita1232534

21.09.2020 17:48

$1) log_3(2x+1)\ \textless \ log_3(x-1)\\3\ \textgreater \ 1 =\ \textgreater \ 2x+1\ \textless \ x-1\\x\ \textless \ -2\\D: \left \{ {{2x+1\ \textgreater \ 0} \atop {x-1\ \textgreater \ 0}} \right. ,\left \{ {{x\ \textgreater \ -\frac{1}{2}} \atop {x\ \textgreater \ 1}} \right. , x\ \textgreater \ 1\\ \\ \left \{ {{x\ \textless \ -2} \atop {x\ \textgreater \ 1}} \right.$
нет решений

$2) log_{\frac{1}{3}}(2-x)\ \textgreater \ 0\\log_{\frac{1}{3}}(2-x)\ \textgreater \ log_{\frac{1}{3}}1\\\frac{1}{3}\ \textless \ 1 =\ \textgreater \ 2-x\ \textless \ 1\\x\ \textgreater \ 1\\ \\D: 2-x\ \textgreater \ 0, x\ \textless \ 2\\ \\ \left \{ {{x\ \textgreater \ 1} \atop {x\ \textless \ 2}} \right. 1\ \textless \ x\ \textless \ 2$

$3) log_{31}(31x+2)\ \textless \ 1\\log_{31}(31x+2)\ \textless \ log_{31}31\\31\ \textgreater \ 1 =\ \textgreater \ 31x+2\ \textgreater \ 31\\31x\ \textgreater \ 29\\x\ \textgreater \ \frac{29}{31}\\ \\ D: 31x+2\ \textgreater \ 0, 31x\ \textgreater \ -2, x\ \textgreater \ -\frac{2}{31}\\ \\ \left \{ {{x\ \textgreater \ \frac{29}{31}} \atop {x\ \textgreater \ -\frac{2}{31}}} \right. x\ \textgreater \ \frac{29}{31}$

$4) log_{\frac{1}{11}}(2x-1)+log_{\frac{1}{11}}x\ \textgreater \ 0\\log_{\frac{1}{11}}(2x-1)\ \textgreater \ -log_{\frac{1}{11}}x\\log_{\frac{1}{11}}(2x-1)\ \textgreater \ log_{\frac{1}{11}}x^{-1}\\log_{\frac{1}{11}}(2x-1)\ \textgreater \ log_{\frac{1}{11}}\frac{1}{x}\\\frac{1}{11}\ \textless \ 1=\ \textgreater \ 2x-1\ \textless \ \frac{1}{x}\\\frac{2x^2-x-1}{x}\ \textless \ 0\\\frac{2(x-1)(x+\frac{1}{2})}{x}\ \textless \ 0\\x\ \textless \ -\frac{1}{2}, x\ \textgreater \ 1$
$D: \left \{ {{2x-1\ \textgreater \ 0} \atop {x\ \textgreater \ 0}} \right. \\ \left \{ {{x\ \textgreater \ \frac{1}{2}} \atop {x\ \textgreater \ 0}} \right. ,x\ \textgreater \ \frac{1}{2}\\ \\ \left \{ {{x\ \textless \ -\frac{1}{2}, x\ \textgreater \ 1} \atop {x\ \textgreater \ \frac{1}{2}}} \right. , x\ \textgreater \ 1$