\(\sqrt[3]{{\dfrac{{4x}}{{2x-3}}}} + 6\sqrt[3]{{\dfrac{{2x-3}}{{4x}}}} = 5.\)
Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{2x-3 \ne 0,}\\{4x \ne 0\;\;\;\;\;\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x \ne 1,5,}\\{x \ne 0\;\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;x \in \left( {-\infty ;0} \right) \cup \left( {0;1,5} \right) \cup \left( {1,5;\infty } \right).\)
Пусть \(\sqrt[3]{{\dfrac{{4x}}{{2x-3}}}} = t.\) Тогда исходное уравнение примет вид:
\(t + \dfrac{6}{t} = 5\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{t \ne 0,\;\;\;\;\;\;\;\;\;\;\;\,\,}\\{{t^2}-5t + 6 = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{t \ne 0,\;\,\,}\\{\left[ {\begin{array}{*{20}{c}}{{t} = 2,}\\{{t} = 3\,}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{t} = 2,}\\{{t} = 3.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{\sqrt[3]{{\dfrac{{4x}}{{2x-3}}}} = 2,}\\{\sqrt[3]{{\dfrac{{4x}}{{2x-3}}}} = 3\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\dfrac{{4x}}{{2x-3}} = 8,\;}\\{\dfrac{{4x}}{{2x-3}} = 27}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\dfrac{{4x-16x + 24}}{{2x-3}} = 0,}\\{\dfrac{{4x-54x + 81}}{{2x-3}} = 0\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{-12x + 24 = 0,}\\{-50x + 81 = 0\;\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 2,\,\,\;}\\{x = \dfrac{{81}}{{50}}.}\end{array}} \right.\)
ОТВЕТ: 81/50; 2.