\(\dfrac{{{x^2}}}{3} + \dfrac{{48}}{{{x^2}}} = 10\,\left( {\dfrac{x}{3}-\dfrac{4}{x}} \right).\)
Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x \ne 0,\,}\\{{x^2} \ne 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;x \ne 0.\)
Пусть \(\dfrac{x}{3}-\frac{4}{x} = t.\) Тогда:
\({\left( {\dfrac{x}{3}-\dfrac{4}{x}} \right)^2} = {t^2}\;\;\;\; \Leftrightarrow \;\;\;\;\dfrac{{{x^2}}}{9}-2 \cdot \dfrac{x}{3} \cdot \dfrac{4}{x} + \dfrac{{16}}{{{x^2}}} = {t^2}\;\;\;\; \Leftrightarrow \;\;\;\;\dfrac{{{x^2}}}{9} + \dfrac{{16}}{{{x^2}}} = {t^2} + \dfrac{8}{3}\left| { \cdot 3} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\dfrac{{{x^2}}}{3} + \dfrac{{48}}{{{x^2}}} = 3{t^2} + 8.\)
Полученное уравнение примет вид:
\(3{t^2} + 8 = 10t\;\;\;\; \Leftrightarrow \;\;\;\;3{t^2}-10t + 8 = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{t = 2,\,}\\{t = \dfrac{4}{3}.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{\dfrac{x}{3}-\dfrac{4}{x} = 2,}\\{\dfrac{x}{3}-\dfrac{4}{x} = \dfrac{4}{3}\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2}-6x-12 = 0,}\\{{x^2}-4x-12 = 0\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 3 \pm \sqrt {21} ,}\\{\begin{array}{*{20}{c}}{x = 6,\;\;\;\,\,\,\;\;\;\;\;\,}\\{x = -2.\;\;\,\;\;\;\;\;\,}\end{array}}\end{array}} \right.\)
Ответ: \(-2;\;\;\;\;6;\;\;\;\;3 \pm \sqrt {21} .\)