\(\dfrac{{{{\left( {x-2} \right)}^2}}}{2} + \dfrac{{18}}{{{{\left( {x-2} \right)}^2}}} = 7\left( {\dfrac{{x-2}}{2}-\dfrac{3}{{x-2}}} \right) + 10.\)
Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{{{\left( {x-2} \right)}^2} \ne 0,}\\{x-2 \ne 0\;\;\;\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \,\;\;\;x \ne 2.\)
Пусть \(\dfrac{{x-2}}{2}-\dfrac{3}{{x-2}} = t.\) Тогда:
\({\left( {\dfrac{{x-2}}{2}-\dfrac{3}{{x-2}}} \right)^2} = {t^2}\;\;\;\, \Leftrightarrow \;\;\;\,\;\dfrac{{{{\left( {x-2} \right)}^2}}}{4}-2 \cdot \dfrac{{x-2}}{2} \cdot \dfrac{3}{{x-2}} + \dfrac{9}{{{{\left( {x-2} \right)}^2}}} = {t^2}\,\,{\rm{|}} \cdot {\rm{2}}\;\;{\rm{ }} \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\,\,\,\dfrac{{{{\left( {x-2} \right)}^2}}}{2} + \dfrac{18}{{{{\left( {x-2} \right)}^2}}} = 2{t^2} + 6.\)
Исходное уравнение примет вид:
\(2{t^2} + 6 = 7t + 10\;\;\,\,\, \Leftrightarrow \;\,\;\;2{t^2}-7t-4 = 0\;\;\;\; \Leftrightarrow \,\;\;\;\left[ {\begin{array}{*{20}{c}}{t = 4,\;\;\;}\\{t = -\dfrac{1}{2}.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{\dfrac{{x-2}}{2}-\dfrac{3}{{x-2}} = 4,\;\;\;}\\{\;\dfrac{{x-2}}{2}-\dfrac{3}{{x-2}} = -\dfrac{1}{2}}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\dfrac{{{x^2}-4x + 4-6-8x + 16}}{{2\left( {x-2} \right)}} = 0,}\\{\dfrac{{{x^2}-4x + 4-6 + x-2}}{{2\left( {x-2} \right)}} = 0\,\;\;\;\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2}-12x + 14 = 0,}\\{{x^2}-3x-4 = 0\,\;\;\,\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 6 \pm \sqrt {22} ,}\\{\begin{array}{*{20}{c}}{x = -1,\;\;\;\;\;\;\;\,}\\{x = 4.\;\;\;\;\;\;\;\;\;\,}\end{array}}\end{array}} \right.\)
Ответ: \(-1;\;\;\;4;\;\;\;6 \pm \sqrt {22} .\)