Log0,5(x^2-7x+12)>log0,5(17-3x)

Question 1

Question 2

$log_{0.5} ( x^{2} -7x+12)\ \textgreater \ log_{0.5}(17-3x)$

ОДЗ:

$\left \{ {{ x^{2} -7x+12\ \textgreater \ 0} \atop {17-3x\ \textgreater \ 0}} \right.$

$\left \{ {{(x-3)(x-4)\ \textgreater \ 0} \atop {-3x\ \textgreater \ -17}} \right.$

$\left \{ {{(x-3)(x-4)\ \textgreater \ 0} \atop {x\ \textless \ 5 \frac{2}{3} }} \right.$

$x^2-7x+12=0$
$D=(-7)^2-4*1*12=49-48=1$
$x_1= \frac{7+1}{2}=4$
$x_2= \frac{7-1}{2}=3$
$x^{2} -7x+12=(x-3)(x-4)$

+ - +
---------------(3)----------(4)----------------
//////////////// ///////////////
-----------------------------------(5 2/3)-----
//////////////////////////////////////

$x$ ∈ $(-$ ∞ $;3)$ ∪ $(4;5 \frac{2}{3} )$

$x^{2} -7x+12\ \textless \ 17-3x$

$x^{2} -7x+12-17+3x\ \textless \ 0$

$x^{2} -4x-5\ \textless \ 0$

$D=(-4)^2-4*1*(-5)=16+20=36$

$x_1= \frac{4+6}{2}=5$

$x_2= \frac{4-6}{2} =-1$

+ - +
----------( -1)------------(5)------------
///////////////

С учётом ОДЗ получаем

Ответ: $(-1;3)$ ∪ $(4;5)$

mukus13_zn · Answer 1 · 2018-03-14T20:23:08+0000

$log_{0.5} ( x^{2} -7x+12)\ \textgreater \ log_{0.5}(17-3x)$

ОДЗ:

$\left \{ {{ x^{2} -7x+12\ \textgreater \ 0} \atop {17-3x\ \textgreater \ 0}} \right.$

$\left \{ {{(x-3)(x-4)\ \textgreater \ 0} \atop {-3x\ \textgreater \ -17}} \right.$

$\left \{ {{(x-3)(x-4)\ \textgreater \ 0} \atop {x\ \textless \ 5 \frac{2}{3} }} \right.$

$x^2-7x+12=0$
$D=(-7)^2-4*1*12=49-48=1$
$x_1= \frac{7+1}{2}=4$
$x_2= \frac{7-1}{2}=3$
$x^{2} -7x+12=(x-3)(x-4)$

+ - +
---------------(3)----------(4)----------------
//////////////// ///////////////
-----------------------------------(5 2/3)-----
//////////////////////////////////////

$x$ ∈ $(-$ ∞ $;3)$ ∪ $(4;5 \frac{2}{3} )$

$x^{2} -7x+12\ \textless \ 17-3x$

$x^{2} -7x+12-17+3x\ \textless \ 0$

$x^{2} -4x-5\ \textless \ 0$

$D=(-4)^2-4*1*(-5)=16+20=36$

$x_1= \frac{4+6}{2}=5$

$x_2= \frac{4-6}{2} =-1$

+ - +
----------( -1)------------(5)------------
///////////////

С учётом ОДЗ получаем

Ответ: $(-1;3)$ ∪ $(4;5)$