\(\dfrac{{{{\left( {x-1} \right)}^2}}}{8} + \dfrac{8}{{{{\left( {x-1} \right)}^2}}} = 7\left( {\dfrac{{x-1}}{4}-\dfrac{2}{{x-1}}} \right)-1.\)
Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{{{\left( {x-1} \right)}^2} \ne 0,}\\{x-1 \ne 0\;\;\;\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \,\;\;\;x \ne 1.\)
Пусть \(\dfrac{{x-1}}{4}-\dfrac{2}{{x-1}} = t.\) Тогда:
\({\left( {\dfrac{{x-1}}{4}-\dfrac{2}{{x-1}}} \right)^2} = {t^2}\;\;\;\, \Leftrightarrow \;\;\;\,\;\dfrac{{{{\left( {x-1} \right)}^2}}}{{16}}-2 \cdot \dfrac{{x-1}}{4} \cdot \dfrac{2}{{x-1}} + \dfrac{4}{{{{\left( {x-1} \right)}^2}}} = {t^2}\,\,{\rm{|}} \cdot {\rm{2}}\;\;{\rm{ }} \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\,\,\,\dfrac{{{{\left( {x-1} \right)}^2}}}{8} + \dfrac{8}{{{{\left( {x-1} \right)}^2}}} = 2{t^2} + 2.\)
Исходное уравнение примет вид:
\(2{t^2} + 2 = 7t-1\;\;\,\,\, \Leftrightarrow \;\,\;\;2{t^2}-7t + 3 = 0\;\;\;\; \Leftrightarrow \,\;\;\;\left[ {\begin{array}{*{20}{c}}{t = 3,\,}\\{t = \dfrac{1}{2}.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{\dfrac{{x-1}}{4}-\dfrac{2}{{x-1}} = 3,\;}\\{\;\dfrac{{x-1}}{4}-\dfrac{2}{{x-1}} = \dfrac{1}{2}}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\dfrac{{{x^2}-2x + 1-8-12x + 12}}{{4\left( {x-1} \right)}} = 0,}\\{\dfrac{{{x^2}-2x + 1-8-2x + 2}}{{4\left( {x-1} \right)}} = 0\,\;\;\;\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2}-14x + 5 = 0,}\\{{x^2}-4x-5 = 0\,\;\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 7 \pm 2\sqrt {11} ,}\\{\begin{array}{*{20}{c}}{x = -1,\;\;\;\;\;\;\;\;\;\,}\\{x = 5.\;\;\;\;\;\;\;\;\;\;\;\,}\end{array}}\end{array}} \right.\)
Ответ: \(-1;\;\;\;5;\;\;\;7 \pm 2\sqrt {11} .\)