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

ОДЗ: + - +---------------(3)----------(4)----------------//////////////// ///////////////-----------------------------------(5 2/3)-----////////////////////////////////////// ∈ ∞ ∪ + - + ----------( -1)------------(5)------------ ///////////////С учётом ОДЗ получаемответ: ∪ ...

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

Ответы

kalabukhovalina 28.06.2020 16:22

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

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

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

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

∈

∞

∪ $(4;5 \frac{2}{3} )$

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

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

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

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)