Question

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Answer 1

$\displaystyle \sf \lg (x-1)+\lg (x+1)=3\lg 2+\lg (x-2)\\\\ ODZ:\begin{cases} x-2>0 \\ x+1>0 \\ x-1>0 \end{cases} \Rightarrow x>2\\\\\lg ((x-1)(x+1))=\lg 2^3+\lg (x-2)\\\\ \lg ((x-1)(x+1))=\lg (8(x-2))\\\\y=\lg t \\\\ (x-1)(x+1)=8(x-2)\\\\x^2-1-8x+16=0\\\\ x^2-8x+15=0\\\\ \left \{ {{x_1x_2=15} \atop {x_1+x_2=8}} \right. \left [{ {{x=5} \atop {x=3}} \right.$

Ответ: $3;5$

$\sf\displaystyle \log_{0,5}\log_6\frac{x^2+x}{x+4} \leq 0\\ ODZ: \begin{cases} \frac{x^2+x}{x+4}>0\\ \log_6\frac{x^2+x}{x+4}>0\end{cases} \Leftrightarrow \; \begin{cases} \frac{x(x+1)}{x+4}>0 \;\;\;\;\;[1]\\ \frac{x^2+x}{x+4}>1\;\;\;\;\;\;\;[2]\end{cases}\\ \;[1]\\ \frac{x(x+1)}{x+4}>0; \; \frac{x(x+1)}{x+4}=0; \\ x=0;\;x=-1;\; x\ne -4\\---(-4)+++(-1)---(0)+++>x$

$\displaystyle \sf \;[2]\\ \frac{x^2+x}{x+4}>1; \; \frac{x^2+x-x-4}{x+4}>0; \; \frac{x^2-4}{x+4}>0;\; \frac{(x-2)(x+2)}{x+4}>0\\x\ne -4;\; x=-2;\; x=2\\ ---(-4)+++(-2)---(2)+++>x\\\\ \begin{cases} ---(-4)+++(-1)---(0)+++>x\\ ---(-4)+(-2)-------(2)+>x \end{cases} \Rightarrow x\in(-4;-2)\cup (2;+\infty)$

$\displaystyle \sf \log_{0,5} \log_6\frac{x^2+x}{x+4}\leq 0 \\\\ y=\log_{0,5}t\\\\ \log_6\frac{x(x+1)}{x+4}\geq 1 \\\\ y=\log_6t\\\\ \frac{x(x+1)}{x+4}\geq 6;\;\;\;\;\; \frac{x^2+x-6x-24}{x+4}\geq 0;\;\;\;\; \frac{x^2-5x-24}{x+4}\geq 0 \\\\ x^2-5x-24=0\\ \left \{ {{x_1x_2=-24} \atop {x_1+x_2=5}} \right. \left [{ {{x=8} \atop {x=-3}} \right. \\\\ \frac{x^2-5x-24}{x+4}\geq 0;\;\;\;\; x=8;\; x=-3;\; x\ne -4;\\\\ ---(-4)+++[-3]---[8]+++>x$

С учётом ОДЗ: $\left \{ {{---(-4)++[-3]-----[8]+>x} \atop {---(-4)++++(-2)-(2)+++>x} \right. \Rightarrow x\in (-4;-3]\cup [8;+\infty)$

Ответ: $(-4;-3]\cup [8;+\infty)$