Задача 11А. Решите неравенство \(\left| {\frac{{3x + 1}}{{x-3}}} \right| < 3\)
ОТВЕТ: \(\left( {-\infty ;\;\frac{4}{3}} \right).\)
\(\left| {\frac{{3x + 1}}{{x-3}}} \right| < 3.\) Неравенство вида \(\left| {f\left( x \right)} \right| < a\) при \(a > 0\) равносильно двойному неравенству: \(-a < f\left( x \right) < a\;\;\;\; \Leftrightarrow \;\;\;\;\,\left\{ {\begin{array}{*{20}{c}}{f\left( x \right) < a,\;\,}\\{f\left( x \right) > -a.}\end{array}} \right.\) \(\left| {\frac{{3x + 1}}{{x-3}}} \right| < 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-3 < \frac{{3x + 1}}{{x-3}} < 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\frac{{3x + 1}}{{x-3}} < 3,\,\,}\\{\frac{{3x + 1}}{{x-3}} > -3}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\frac{{3x + 1}}{{x-3}}-3 < 0,\,\,}\\{\frac{{3x + 1}}{{x-3}} + 3 > 0\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\frac{{10}}{{x-3}} < 0,\,}\\{\frac{{6x-8}}{{x-3}} > 0}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \in \left( {-\infty ;\,3} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \in \left( {-\infty ;\frac{4}{3}} \right) \cup \left( {3;\, + \infty } \right)}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;\frac{4}{3}} \right).\) Ответ: \(\left( {-\infty ;\;\frac{4}{3}} \right).\)