Профиль №15. Неравенства с модулями. Задача 15А

ОТВЕТ: \(\left( {3;\;\infty } \right).\)

\(\left| {x-1} \right| < 2x-4.\)

Неравенство вида  \(\left| {f\left( x \right)} \right| < g\left( x \right)\)  равносильно двойному неравенству:

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

\(\left| {x-1} \right| < 2x-4\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-2x + 4 < x-1 < 2x-4\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x-1 < 2x-4\,\,\,}\\{x-1 > -2x + 4}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x > 3,}\\{x > \dfrac{5}{3}}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x \in \left( {3;\, + \infty } \right).\)

Ответ:  \(\left( {3; + \infty } \right)\).