tex]...

September 2021 1 47 Report

Решить неравенство

[tex]\log_{x^2-x+1}(x^2+6x+8)+\log_{\frac{1}{x^2-x+1}}(x+2)\ \textless \ \frac{1}{\log_{5/2}(x^2-x+1)}[/tex]

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mukus13

Verified answer

$\log_{x^2-x+1}(x^2+6x+8)+\log_{\frac{1}{x^2-x+1}}(x+2)\ \textless \ \frac{1}{\log_{ \frac{5}{2} }(x^2-x+1)}$

Найдем ОДЗ:

$1)$
$x^2-x+1 \neq 1}$

${{x^2-x \neq 0}$

$x(x-1) \neq 0$

$x \neq 0,$ $x \neq 1$

$2)$
$x^{2} -x+1\ \textgreater \ 0$

$x^{2} -x+1=0$

$D=(-1)^2-4*1*1\ \textless \ 0$

$a=1\ \textgreater \ 0$

парабола $y=x^2-x+1$ расположена над осью OX, т. е. $y\ \textgreater \ 0$ при любом значении x, поэтому данное неравенство верно при любых значениях x, т. е. $x$ ∈ $R$

$3)$
$x^{2} +6x+8\ \textgreater \ 0$

$x^2+6x+8=0$

$D=6^2-4*1*8=4$

$x_1= \frac{-6+2}{2} =-2$

$x_2= \frac{-6-2}{2} =-4$

$(x+2)(x+4)\ \textgreater \ 0$

$4)$
$log_{2.5}(x^2-x+1) \neq 0$

$log_{2.5}(x^2-x+1) \neq log_{2.5}1$

$x^2-x+1 \neq 1$

$x \neq 0,$ $x \neq 1$

$5)$
$x+2\ \textgreater \ 0$

$x\ \textgreater \ -2$

Отметим на числовой прямой и найдем общее решение ОДЗ:

+ - +
----------(-4)---------------(-2)-------------------
////////////// /////////////////
------------------------------------(0)-----(1)---------
/////////////////////////////////////////////////////////
-------------------------------(-2)---------------------
/////////////////

$x$ ∈ $(-2;0)$ ∪ $(0;1)$ ∪ $(1;+$ ∞ $)$

$\log_{x^2-x+1}(x^2+6x+8)+\log_{{(x^2-x+1)^{-1}}}(x+2)\ \textless \ {\log_{x^2-x+1}2.5}$

$\log_{x^2-x+1}(x^2+6x+8)-\log_{{x^2-x+1}}(x+2)\ \textless \ {\log_{x^2-x+1}2.5}$

$\log_{x^2-x+1}(x^2+6x+8)\ \textless \ {\log_{x^2-x+1}2.5}+\log_{{x^2-x+1}}(x+2)$

$\log_{x^2-x+1}(x^2+6x+8)\ \textless \ {\log_{x^2-x+1}[2.5(x+2)]}$

$\log_{x^2-x+1}(x^2+6x+8)\ \textless \ {\log_{x^2-x+1}(2.5x+5)}$

$1)$

$\left \{ {{0\ \textless \ x^2-x+1\ \textless \ 1} \atop {x^2+6x+8\ \textgreater \ 2.5x+5}}} \right.$

$\left \{ {{x^2-x+1\ \textgreater \ 0}\atop {x^2-x+1\ \textless \ 1}} \atop {x^2+6x+8\ \textgreater \ 2.5x+5}}} \right.$

$\left \{ {{x^2-x+1\ \textgreater \ 0}\atop {x^2-x\ \textless \ 0}} \atop {x^2+3.5x+3\ \textgreater \ 0}}} \right.$

$\left \{ {{x^2-x+1\ \textgreater \ 0}\atop {x(x-1)\ \textless \ 0}} \atop {x^2+3.5x+3\ \textgreater \ 0}}} \right.$

$x^{2} +3.5x+3=0$
$D=3.5^2-4*1*3=12.25-12=0.25$
$x_1= \frac{-3.5+0.5}{2}=-1.5$
$x_2= \frac{-3.5-0.5}{2}=-2$
$x^{2} -x+1\ \textgreater \ 0$ ⇒ $x$ ∈ $R$

+ - +
-------------------------(0)-------(1)-------------
///////////
+ - +
-----(-2)-----(-1.5)-------------------------------
//////// //////////////////////////////

$x$ ∈ $(0;1)$

$2)$

$\left \{ {{x^2-x+1\ \textgreater \ 1} \atop {x^2+6x+8\ \textless \ 2.5x+5}}} \right.$

$\left \{ {{x^2-x\ \textgreater \ 0} \atop {x^2+3.5x+3\ \textless \ 0}}} \right.$

$\left \{ {{x(x-1)\ \textgreater \ 0} \atop {(x+1.5)(x+2)\ \textless \ 0}}} \right.$

+ - +
-----------------------------(0)-------------(1)--------------
/////////////////////////////////// //////////////
+ - +
---------(-2)-------(-1.5)------------------------------
//////////

$x$ ∈ $(-2;-1.5)$

Объединяем 1) и 2) случаи и получаем общее решение неравенства:

$x$ ∈ $(-2;-1.5)$ ∪ $(0;1)$

Найдем пересечение с ОДЗ:

-------------------------(-2)--------------(0)-----(1)-----------------
/////////////////////////////////////////////
--------------------------(-2)---(-1.5)----(0)-----(1)----------------
/////// ////////

Ответ: $(-2;-1.5)$ ∪ $(0;1)$

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