Question

Решите неравенство умоляю пожалуйста
Лучше с полным решением

Answer 1

$ODZ:x>0\\\\log_{2}^{4}x-log_{0,5}^{2} \frac{x^{3} }{8}+9log_{2}\frac{32}{x^{2}} <4log_{0,5}^{2}x\\\\log_{2}^{4}x-(log_{0,5} x^{3}+log_{0,5} \frac{1}{8})^{2} +9(log_{2}32-log_{2}x^{2})<4log_{2} ^{2}x\\\\log_{2}^{4}x-(-3log_{2} x+3)^{2}+9(5-2log_{2}x)<4log_{2}^{2}x\\\\log_{2}^{4}x-9+18log_{2} x-9log_{2}^{2}x+45-18log_{2}x-4log_{2}^{2}x<0 \\\\log_{2}^{4}x-13log_{2} ^{2}x+36<0\\\\log_{2} ^{2} x=m\\\\m^{2} -13m+36<0\\\\(m-9)(m-4)<0$

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            4             9

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m ∈ (4 ; 9)

$1)log_{2}^{2}x>4\\\\(log_{2}x-2)(log_{2}x+2)>0\\\\\left \{ {{log_{2}x<-2 } \atop {log_{2}x>2 }} \right. \\\\\left \{ {{x<0,25} \atop {x>4}} \right. \ \Rightarrow (\ x\in{0 \ ; \ 0,25)\cup(4 \ ; \ +\infty)$

$2)log_{2}^{2}x<9\\\\(log_{2} x-3)(log_{2} x+3)<0\\\\\left \{ {{log_2}x>-3 } \atop {log_{2}x<3 }} \right.\\\\\left \{ {{x>0,125} \atop {x<8}} \right. \ \Rightarrow x\in(0,125 \ ; \ 8)\\\\\boxed{x\in(0,125 \ ; \ 0,25)\cup(4 \ ; \ 8)}$