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