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$\left \{ {{5^{3x-1}-5^{3x+1} \leq -72} \atop {log_{ \frac{x}{3}} (3x^2-2x+1) \geq 0}}} \right.$
ОДЗ:
$\left \{ {{3x^2-2x+1\ \textgreater \ 0} \atop { \frac{x}{3}\ \textgreater \ 0 }} \atop { \frac{x}{3} \neq 1}} \right.$
$\left \{ {{3x^2-2x+1\ \textgreater \ 0} \atop { x\ \textgreater \ 0 }} \atop {x \neq 3}} \right.$
$3x^2-2x+1=0$
$D=(-2)^2-4*3*1\ \textless \ 0$
$x$ ∈ $R$

------------(0)------------(3)----------------
////////////////////////////////////
$x$ ∈ $(0;3)$ ∪ $(3;+$ ∞ $)$

Решим отдельно первое неравенство:
$5^{3x-1}-5^{3x+1} \leq -72$
$5^{3x}* \frac{1}{5} -5^{3x}*5 \leq -72$
$5^{3x}( \frac{1}{5}- 5) \leq -72$
$-4,8*5^{3x} \leq -72$
$5^{3x} \geq 15$
$log_5{5^{3x}} \geq log_515$
${3x}} \geq log_515$
$x \geq \frac{log_515}{3}$
$x \geq \frac{log_515}{log_5125}$
$x \geq {log_{125}15}$
$[log_{125}15;+$ ∞ $)$
Решим отдельно второе неравенство:
$log_{ \frac{x}{3}} (3x^2-2x+1) \geq 0$
$log_{ \frac{x}{3}} (3x^2-2x+1) \geq log_{ \frac{x}{3} }1$
1)
$\left \{ {{0\ \textless \ \frac{x}{3}\ \textless \ 1} \atop {3x^2-2x+1\leq }1}} \right.$
$\left \{ {{0\ \textless \ x\ \textless \ 3} \atop {3x^2-2x+1-1\leq }0}} \right.$
$\left \{ {{0\ \textless \ x\ \textless \ 3} \atop {3x^2-2x\leq }0}} \right.$
$\left \{ {{0\ \textless \ x\ \textless \ 3} \atop {x(3x-2)\leq }0}} \right.$

--------(0)-----------------(1)------------
//////////////////
--------[0]--------[2/3]--------------------
/////////////
$x$ ∈ $(0; \frac{2}{3}]$
2)
$\left \{ {{ \frac{x}{3}\ \textgreater \ 1 } \atop {3x^2-2x+1 \geq 1}} \right.$
$\left \{ {{x\ \textgreater \ 3} \atop {3x^2-2x \geq 0}} \right.$
$\left \{ {{x\ \textgreater \ 3} \atop {x(3x-2) \geq 0}} \right.$

---------------------------(3)------------------
///////////////////
--------[0]-------[2/3]------------------------
/////////// ////////////////////////
$x$ ∈ $(3;+$ ∞ $)$

тогда общее для 1) и 2) случая логарифмического неравенства
$x$ ∈ $(0; \frac{2}{3}]$ ∪ $(3;+$ ∞ $)$

найдем общее решение показательного и логарифмического неравенств:

------------(0)----------------------[2/3]-------------(3)------------------
//////////////////////////// ////////////////
------------------[log_{125} 15]------------------------------------------
/////////////////////////////////////////////////////////////
ОДЗ: ----(0)-----------------------------------------(3)------------------
////////////////////////////////////////////////////////////////

Ответ: $[log_{125}15; \frac{2}{3}]$ ∪ $(3;+$ ∞ $)$

Luluput_zn (192k баллов) 14 Май, 18