Задача 53. Решите неравенство \(\dfrac{{\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{\left| {x-1} \right|}} \ge \dfrac{{\lg {{\left( {-{x^2}-2x + 2} \right)}^2}}}{{x-1}}\).
ОТВЕТ: \(\left( {-\infty ;\;-3} \right] \cup \left( {1;\;\infty } \right).\)
\(\dfrac{{\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{\left| {x-1} \right|}} \ge \dfrac{{\lg {{\left( {-{x^2}-2x + 2} \right)}^2}}}{{x-1}}.\) Найдём ограничения на подлогарифмические выражения: \(\left\{ {\begin{array}{*{20}{c}}{{x^2} + 2x-2 > 0,\,\,\,\,\,\,\,\,\,}\\{{{\left( {-{x^2}-2x + 2} \right)}^2} > 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;{x^2} + 2x-2 > 0\;\;\;\; \Leftrightarrow \;\;\;\;x\, \in \,\left( {-\infty ;-1-\sqrt 3 } \right) \cup \left( {-1 + \sqrt 3 ;\infty } \right).\) Вернёмся к исходному неравенству: \(\dfrac{{\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{\left| {x-1} \right|}} \ge \dfrac{{\lg {{\left( {-{x^2}-2x + 2} \right)}^2}}}{{x-1}}\;\;\;\; \Leftrightarrow \;\;\;\;\dfrac{{\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{\left| {x-1} \right|}} \ge \dfrac{{2\lg \left| {-{x^2}-2x + 2} \right|}}{{x-1}}.\) Так как \({x^2} + 2x-2 > 0,\) то \(\left| {-{x^2}-2x + 2} \right| = \left| {{x^2} + 2x-2} \right| = {x^2} + 2x-2.\) Тогда полученное неравенство примет вид: \(\dfrac{{\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{\left| {x-1} \right|}} \ge \dfrac{{2\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}}\;\;\;\; \Leftrightarrow \) \( \,\,\,\,\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x > 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}}-\dfrac{{2\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}} \ge 0\;\,\,\;}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x < 1,\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{-\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}}-\dfrac{{2\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}} \ge 0.}\end{array}} \right.}\end{array}} \right.\) Рассмотрим первую систему: \(\left\{ {\begin{array}{*{20}{c}}{x > 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\,}\\{\dfrac{{\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}}-\dfrac{{2\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}} \ge 0.}\end{array}} \right.\) \(\dfrac{{\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}}-\dfrac{{2\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}} \ge 0\;\;\;\; \Leftrightarrow \;\;\;\;\;\dfrac{{\left( {{x^2} + x-2} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}} \ge 0.\) Решим полученное неравенство методом интервалов при условии, что \(x > 1.\) Найдём нули числителя: \(\left[ {\begin{array}{*{20}{c}}{{x^2} + x-2 = 0,\;\;\;\;\;\;\;}\\{\lg \left( {{x^2} + 2x-2} \right) = 0}\end{array}\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2} + x-2 = 0,\;\;\;\;\;\;\;\;\,\;}\\{\lg \left( {{x^2} + 2x-2} \right) = \lg 1}\end{array}\;\;\;\; \Leftrightarrow } \right.} \right.\) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2} + x-2 = 0,\;}\\{{x^2} + 2x-3 = 0}\end{array}\;\;\;\; \Leftrightarrow \;\;\;\;} \right.\left[ {\begin{array}{*{20}{c}}{\left( {x-1} \right)\left( {x + 2} \right) = 0,}\\{\left( {x-1} \right)\left( {x + 3} \right) = 0\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -3,}\\{x = -2,}\\{x = 1.\;\;\,}\end{array}} \right.\) Найдём нули знаменателя: \(x = 1.\) Следовательно, решение первой системы: \(x\, \in \,\left( {1;\infty } \right).\) Рассмотрим вторую систему: \(\left\{ {\begin{array}{*{20}{c}}{x < 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{-\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}}-\dfrac{{2\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}} \ge 0.}\end{array}} \right.\) \(\dfrac{{-\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}}-\dfrac{{2\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}} \ge 0\;\;\;\; \Leftrightarrow \;\;\;\;\dfrac{{\left( {-{x^2}-x-2} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}} \ge 0.\) Решим полученное неравенство методом интервалов при условии, что \(x < 1.\) Найдем нули числителя: \(\left[ {\begin{array}{*{20}{c}}{-{x^2}-x-2 = 0,\;\;\;\;\;}\\{\lg \left( {{x^2} + 2x-2} \right) = 0}\end{array}\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2} + x + 2 = 0,\;\;\;\;\;\;\;\;\,\;}\\{\lg \left( {{x^2} + 2x-2} \right) = \lg 1}\end{array}\;\;\;\; \Leftrightarrow } \right.} \right.\) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2} + x + 2 = 0,\;}\\{{x^2} + 2x-3 = 0}\end{array}\;\;\;\; \Leftrightarrow \;\;\;\;} \right.\left( {x-1} \right)\left( {x + 3} \right) = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -3,}\\{x = 1.\;\;}\end{array}} \right.\) Найдем нули знаменателя: \(x = 1.\) Следовательно, решение второй системы: \(x\, \in \,\left( {-\infty ;-3} \right]\). \(\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x > 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}}-\dfrac{{2\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}} \ge 0\;\,\,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x < 1,\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{-\left( {{x^2} + x} \right)\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}}-\dfrac{{2\lg \left( {{x^2} + 2x-2} \right)}}{{x-1}} \ge 0}\end{array}} \right.}\end{array}} \right.\;\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x \in \left( {1;\infty } \right),\;\;\;\;}\\{x \in \left( {-\infty ;-3} \right].}\end{array}} \right.\) Таким образом, решением исходного неравенства является: \(x \in \left( {-\infty ;\;-3} \right] \cup \left( {1;\;\infty } \right).\) Ответ: \(\left( {-\infty ;\;-3} \right] \cup \left( {1;\;\infty } \right).\) 
