\(\dfrac{{{x^2} + 1}}{x} + \dfrac{x}{{{x^2} + 1}} = -\dfrac{5}{2}.\)
Пусть \(\dfrac{{{x^2} + 1}}{x} = t.\) Тогда исходное уравнение примет вид:
\(t + \dfrac{1}{t} = -\dfrac{5}{2}\;\;\;\; \Leftrightarrow \;\;\;\;\dfrac{{2{t^2} + 5t + 2}}{{2t}} = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{t \ne 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\{2{t^2} + 5t + 2 = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{t \ne 0,\;\,\;\;}\\{\left[ {\begin{array}{*{20}{c}}{t = -2,}\\{t = -\dfrac{1}{2}}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{t = -2,\;}\\{t = -\dfrac{1}{2}.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{\dfrac{{{x^2} + 1}}{x} = -2,}\\{\dfrac{{{x^2} + 1}}{x} = -\dfrac{1}{2}}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\dfrac{{{x^2} + 2x + 1}}{x} = 0,}\\{\dfrac{{2{x^2} + x + 2}}{x} = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x \ne 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\{\left[ {\begin{array}{*{20}{c}}{{x^2} + 2x + 1 = 0,}\\{2{x^2} + x + 2 = 0}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x \ne 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\{\left[ {\begin{array}{*{20}{c}}{{{\left( {x + 1} \right)}^2} = 0,\;\;\;\;\,}\\{2{x^2} + x + 2 = 0}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x \ne 0,\;\;\;}\\{\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x \notin R\;\,}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;x = -1.\)
Ответ: \(-1.\)