Решите неравенство (log2^2 x - 2log2 x )^2 + 36log2 x + 45< 18log2^2 x

Ответ:

190G

24.09.2020 23:45

$\displaystyle (log_2^2x-2log_2x)^2+36log_2x+45\ \textless \ 18log_2^2x$

$\displaystyle (log_2^2x-2log_2x)^2-18(log_2^2x-2log_2x)+45\ \textless \ 0$

$\displaystyle Log_2^2x-2log_2x=t$

$\displaystyle t^2-18t+45\ \textless \ 0$

$\displaystyle D=324-180=144=12^2

t_P1=15; t_2=3$

__+_____-_____+____
3 15

$\displaystyle \left \{ {{Log_2^2x-2log_2x\ \textless \ 15} \atop {log_2^2x-2log^2x\ \textgreater \ 3}} \right.$

1)
$\displaystyle log_2^2x-2log_2x-15\ \textless \ 0$

$\displaystyle D=4+64=64=8^2
$

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-3 5

$\displaystyle -3\ \textless \ log_2x\ \textless \ 5$

$\displaystyle \frac{1}{2^3}\ \textless \ x\ \textless \ 2^5$

2)
$\displaystyle log_2^2x-2log_2x-3\ \textgreater \ 0$

$\displaystyle D=4^2$

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-1 3

$\displaystyle \left \{ {{log_2x\ \textgreater \ 3} \atop {log_2x\ \textless \ -1}} \right.$

$\displaystyle \left \{ {{x\ \textless \ \frac{1}{2}} \atop {x\ \textgreater \ 2^5}} \right.$

Объединим все промежутки с учетом ОДЗ х>0

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0 1/8 1/2 8 32

ОТВЕТ: (1/8; 1/2) ∪(8;32)

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