логаритмични неравенства

[Гост]

Здравейте,
ще съм повече от благодарна, ако някой ми обясни как да реша задачата .Смених основата, положих и не знам как да реша това дето се получи

[ammornil]

[tex]\begin{array}{lcll} \log_{\frac{1}{2}}{x}+\log_{x}{\frac{1}{2}}+3\ge0 && \text{ДМ: } \begin{array}{|l} x>0 \\ x\ne{1} \end{array} \\ && \Rightarrow x \in (0;1)\cup (1:+\infty) \\ \log_{x}{\frac{\normalsize{1}}{\normalsize{2}}}=\frac{ \log_{\frac{1}{2}}{\frac{\normalsize{1}}{\normalsize{2}}}}{ \log_{\frac{1}{2}}{\normalsize{x}}}=\frac{1}{ \log_{\frac{1}{2}}{\normalsize{x}}} \\ & \because & u= \log_{\frac{1}{2}}{\normalsize{x}} \\ && \Rightarrow x=\left(\frac{1}{2}\right)^{u} \\ u+\frac{\normalsize{1}}{\normalsize{u}}+3\ge{0} \\ u^{2}+3u+1\ge{0} && u_{1,2}=\frac{\normalsize{-3\pm\sqrt{3^{2}-4\cdot{1}\cdot{1}}}}{\normalsize{2\cdot{1}}} \\ && \begin{cases} u_{1}=\frac{\normalsize{-3-\sqrt{5}}}{\normalsize{2}} \approx -2,618 \\ u_{2}=\frac{\normalsize{-3+\sqrt{5}}}{\normalsize{2}} \approx -0,382 \end{cases} \\ u \in \left(-\infty;\frac{\normalsize{-3-\sqrt{5}}}{\normalsize{2}}\right] \cup \left[\frac{\normalsize{-3+\sqrt{5}}}{\normalsize{2}};+\infty\right) \\ \Rightarrow x \in \left(\left(\frac{\normalsize{1}}{\normalsize{2}}\right)^{+\infty}; \left(\frac{\normalsize{1}}{\normalsize{2}}\right)^{\frac{\normalsize{-3+\sqrt{5}}}{\normalsize{2}}} \right] \cup \left[\left( \frac{\normalsize{1}}{\normalsize{2}}\right)^{\frac{\normalsize{-3-\sqrt{5}}}{\normalsize{2}}}; \left(\frac{\normalsize{1}}{\normalsize{2}}\right)^{-\infty} \right) \cap \text{ДМ} \\ x \in \left(0; \left(\frac{\normalsize{1}}{\normalsize{2}}\right)^{\frac{\normalsize{-3+\sqrt{5}}}{\normalsize{2}}} \right] \cup \left[\left( \frac{\normalsize{1}}{\normalsize{2}}\right)^{\frac{\normalsize{-3-\sqrt{5}}}{\normalsize{2}}}; +\infty \right)\end{array}[/tex]