Профиль №15. Неравенства с модулями. Задача 9А

ОТВЕТ: \(\left( {-1;\;2} \right) \cup \left( {3;\;6} \right).\) 

\(\left| {{x^2}-5x} \right| < 6.\)

Неравенство вида  \(\left| {f\left( x \right)} \right| < a\)  при  \(a > 0\)  равносильно двойному неравенству:

\(-a < f\left( x \right) < a\;\;\;\; \Leftrightarrow \;\;\;\;\,\left\{ {\begin{array}{*{20}{c}}{f\left( x \right) < a,\;\,}\\{f\left( x \right) > -a.}\end{array}} \right.\)

\(\left| {{x^2}-5x} \right| < 6\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-6 < {x^2}-5x < 6\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}-5x < 6,\,}\\{{x^2}-5x > -6}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}-5x-6 < 0,\,}\\{{x^2}-5x + 6 > 0\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \in \left( {-1;6} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \in \left( {-\infty ;2} \right) \cup \left( {3;\, + \infty } \right)}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-1;\;2} \right) \cup \left( {3;\;6} \right).\)

Ответ:  \(\left( {-1;\;2} \right) \cup \left( {3;\;6} \right).\)