Log0,1(x-2)-lgx>log0,1 3;

Question 1

Question 2

$log_{0,1}(x-2)-lgx\ \textgreater \ log_{0,1} 3$
ОДЗ:
$\left \{ {{x\ \textgreater \ 0} \atop {x-2\ \textgreater \ 0}} \right.$
$\left \{ {{x\ \textgreater \ 0} \atop {x\ \textgreater \ 2}} \right.$

--------------(0)------------------
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-------------------(2)-------------
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$x$ ∈ $(2;+$ ∞ $)$

$log_{10^{-1}}(x-2)-lgx\ \textgreater \ log_{10^{-1}} 3$
$-lg(x-2)-lgx\ \textgreater \ -lg 3$
$lg(x-2)+lgx\ \textless \ \ lg 3$
$lg[x(x-2)]\ \textless \ \ lg 3$
$lg(x^2-2x)\ \textless \ \ lg 3$
$x^2-2x\ \textless \ \ 3$
$x^2-2x-3\ \textless \ 0$
$D=(-2)^2-4*1*(-3)=4+12=16$
$x_1= \frac{2+4}{2} =3$
$x_1= \frac{2-4}{2} =-1$

-----+----(-1)---- - ----(3)------+------
//////////////
-------------------(2)--------------------
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Ответ: (2; 3)

Luluput_zn · Answer 1 · 2018-05-11T16:36:51+0000

$log_{0,1}(x-2)-lgx\ \textgreater \ log_{0,1} 3$
ОДЗ:
$\left \{ {{x\ \textgreater \ 0} \atop {x-2\ \textgreater \ 0}} \right.$
$\left \{ {{x\ \textgreater \ 0} \atop {x\ \textgreater \ 2}} \right.$

--------------(0)------------------
///////////////////
-------------------(2)-------------
////////////
$x$ ∈ $(2;+$ ∞ $)$

$log_{10^{-1}}(x-2)-lgx\ \textgreater \ log_{10^{-1}} 3$
$-lg(x-2)-lgx\ \textgreater \ -lg 3$
$lg(x-2)+lgx\ \textless \ \ lg 3$
$lg[x(x-2)]\ \textless \ \ lg 3$
$lg(x^2-2x)\ \textless \ \ lg 3$
$x^2-2x\ \textless \ \ 3$
$x^2-2x-3\ \textless \ 0$
$D=(-2)^2-4*1*(-3)=4+12=16$
$x_1= \frac{2+4}{2} =3$
$x_1= \frac{2-4}{2} =-1$

-----+----(-1)---- - ----(3)------+------
//////////////
-------------------(2)--------------------
////////////////////

Ответ: (2; 3)