84В. Решите неравенство \(\frac{{\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{\left| {x-2} \right|}} \ge \frac{{{{\log }_8}{{\left( {-{x^2}-4x + 4} \right)}^6}}}{{x-2}}\).
\(\frac{{\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{\left| {x-2} \right|}} \ge \frac{{{{\log }_8}{{\left( {-{x^2}-4x + 4} \right)}^6}}}{{x-2}}.\) Найдём ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{{x^2} + 4x-4 > 0,\,\,\,\,\,\,\,\,\,}\\{{{\left( {-{x^2}-4x + 4} \right)}^6} > 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;{x^2} + 4x-4 > 0\;\;\;\; \Leftrightarrow \;\;\;\;x\, \in \,\left( {-\infty ;-2-2\sqrt 2 } \right) \cup \left( {-2 + 2\sqrt 2 ;\infty } \right).\) Вернёмся к исходному неравенству: \(\frac{{\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{\left| {x-2} \right|}} \ge \frac{{{{\log }_8}{{\left( {-{x^2}-4x + 4} \right)}^6}}}{{x-2}}\;\;\;\; \Leftrightarrow \;\;\;\;\frac{{\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{\left| {x-2} \right|}} \ge \frac{{6{{\log }_8}\left| {-{x^2}-4x + 4} \right|}}{{x-2}}.\) Так как \({x^2} + 4x-4 > 0,\) то \(\left| {-{x^2}-4x + 4} \right| = \left| {{x^2} + 4x-4} \right| = {x^2} + 4x-4.\) Тогда полученное неравенство примет вид: \(\frac{{\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{\left| {x-2} \right|}} \ge \frac{{6{{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}}\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x > 2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\;\;\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}}-\frac{{6{{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}} \ge 0}\end{array}\;\;\;} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x < 2,\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\,\,\,\,\,\,\,\,\,\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{-\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}}-\frac{{6{{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}} \ge 0.}\end{array}} \right.}\end{array}} \right.\) Рассмотрим первую систему: \(\left\{ {\begin{array}{*{20}{c}}{x > 2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}}-\frac{{6{{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}} \ge 0.}\end{array}} \right.\) \(\frac{{\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}}-\frac{{6{{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}} \ge 0\;\;\;\; \Leftrightarrow \;\;\;\;\;\frac{{\left( {{x^2} + x-6} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}} \ge 0.\) Решим полученное неравенство методом интервалов при условии, что \(x > 2.\) Найдём нули числителя: \(\left[ {\begin{array}{*{20}{c}}{{x^2} + x-6 = 0,\;\;\;\;\;\;\;\;\;\;}\\{{{\log }_8}\left( {{x^2} + 4x-4} \right) = 0}\end{array}\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2} + x-6 = 0,\;\;\;\;\;\;\;\;\;\;\;\;\,\;\;\,\;}\\{{{\log }_8}\left( {{x^2} + 4x-4} \right) = {{\log }_8}1}\end{array}\;\;\;\; \Leftrightarrow } \right.} \right.\) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2} + x-6 = 0,\;}\\{{x^2} + 4x-5 = 0}\end{array}\;\;\;\; \Leftrightarrow \;\;\;\;} \right.\left[ {\begin{array}{*{20}{c}}{\left( {x-2} \right)\left( {x + 3} \right) = 0,}\\{\left( {x-1} \right)\left( {x + 5} \right) = 0\,\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 2,\;\;}\\{x = -3,}\\{x = 1,\;\;\,}\\{x = -5.}\end{array}} \right.\) Найдём нули знаменателя: \(x \ne 2.\) Следовательно, решение первой системы: \(x\, \in \,\left( {2;\infty } \right).\) Рассмотрим вторую систему: \(\left\{ {\begin{array}{*{20}{c}}{x < 2,\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\,\,\,\,\,\,\,\,\,\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{-\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}}-\frac{{6{{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}} \ge 0.}\end{array}} \right.\) \(\frac{{-\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}}-\frac{{6{{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}} \ge 0\;\;\;\; \Leftrightarrow \;\;\;\;\frac{{\left( {-{x^2}-x-6} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}} \ge 0.\) Решим полученное неравенство методом интервалов при условии, что \(x < 2.\) Найдем нули числителя: \(\left[ {\begin{array}{*{20}{c}}{-{x^2}-x-6 = 0,\;\;\;\;\;\;\,\;\;}\\{{{\log }_8}\left( {{x^2} + 4x-4} \right) = 0}\end{array}\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{-{x^2}-x-6 = 0,\;\;\;\;\;\;\;\;\,\;\;\;\;\,\;}\\{{{\log }_8}\left( {{x^2} + 4x-4} \right) = {{\log }_8}1}\end{array}\;\;\;\; \Leftrightarrow } \right.} \right.\) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2} + x + 6 = 0,\;}\\{{x^2} + 4x-5 = 0}\end{array}\;\;\;\; \Leftrightarrow \;\;\;\;} \right.\left( {x-1} \right)\left( {x + 5} \right) = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -5,}\\{x = 1.\;\;}\end{array}} \right.\) Найдем нули знаменателя: \(x \ne 2.\) Следовательно, решение второй системы: \(x\, \in \,\left( {-\infty ;-5} \right] \cup \left[ {1;2} \right).\) \( \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x > 2,\,\,\,\,\,\,\,\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}}-\frac{{6{{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}} \ge 0}\end{array}\;\;\;} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x < 2,\;\;\;\,\,\,\;\;\,\,\,\,\,\,\,\,\,\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{-\left( {{x^2} + x} \right){{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}}-\frac{{6{{\log }_8}\left( {{x^2} + 4x-4} \right)}}{{x-2}} \ge 0\,}\end{array}} \right.}\end{array}} \right.\;\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x \in \left( {2;\infty } \right),\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\{x \in \left( {-\infty ;-5} \right] \cup \left[ {1;2} \right).}\end{array}} \right.\) Таким образом, решением исходного неравенства является: \(x \in \left( {-\infty ;\;-5} \right] \cup \left[ {1;\;2} \right) \cup \left( {2;\;\infty } \right).\) Ответ: \(\left( {-\infty ;\;-5} \right] \cup \left[ {1;\;2} \right) \cup \left( {2;\;\infty } \right).\)
