Question

пожалуйста,помогите решить

Answer 1

$\frac{log_2(8x)\cdot log_{0,125x}2}{log_{0,5x}16} \leq \frac{1}{4}\; ,\; \; ODZ:\; \left \{ {{x>0,x\ne 2} \atop {x\ne 8}} \right. \\\\log_2(8x)=log_28+log_2x=3+log_2x\\\\log_{0,125x}2=\frac{1}{log_2{0,125x}}=\frac{1}{log_20,125+log_2x}=\frac{1}{log_22^{-3}+log_2x}=\frac{1}{log_2x-3}\\\\log_{0,5x}16=\frac{1}{log_{16}0,5x}=\frac{1}{log_{2^4}{0,5x}}}=\frac{1}{\frac{1}{4}(log_22^{-1}+log_2x)}=\frac{4}{log_2x-1}\\\\\\\frac{(3+log_2x)(log_2x-1)}{4(log_2x-3)}-\frac{1}{4} \leq 0\\\\t=log_2x,$

$\frac{(t+3)(t-1)}{4(t-3)}-\frac{1}{4} \leq 0\; ,\; \; \to \; \; \frac{(t+3)(t-1)-(t-3)}{4(t-3)} \leq 0\\\\\frac{t^2+2t-3-t+3}{4(t-3)} \leq 0,\; \; \to \; \; \frac{t(t+1)}{4(t-3)} \leq 0\\\\---[-1]+++[0]---(3)+++\\\\t\in (-\infty,-1]U[0,3)\; \; \to \; \; log_2x \leq -1\; \; ili\; \; 0 \leq log_2x<3\\\\x \leq 2^{-1}\; \; ili\; \; \left \{ {{x \geq 2^0} \atop {x<2^3}} \right. \\\\x \leq 0,5\; \; ili\; \; \left \{ {{x \geq 1} \atop {x<8}} \right. ,\\\\Otvet:\; x\in (0;\; 0,5]U[\, 1;2)U(2;8).$