\({x^2} + \dfrac{{25{x^2}}}{{{{\left( {x + 5} \right)}^2}}} = 11\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{x^2}-2 \cdot x \cdot \dfrac{{5x}}{{x + 5}} + {\left( {\dfrac{{5x}}{{x + 5}}} \right)^2} = 11-2 \cdot x \cdot \dfrac{{5x}}{{x + 5}}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\left( {x-\dfrac{{5x}}{{x + 5}}} \right)^2} = 11-10\dfrac{{{x^2}}}{{x + 5}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {\dfrac{{{x^2}}}{{x + 5}}} \right)^2} + 10\dfrac{{{x^2}}}{{x + 5}}-11 = 0.\)
Пусть \(\dfrac{{{x^2}}}{{x + 5}} = t\). Тогда:
\({t^2} + 10t-11 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,\,\,\,\,\,}\\{t = -11.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{\dfrac{{{x^2}}}{{x + 5}} = 1\,\,\,\,\,\,}\\{\dfrac{{{x^2}}}{{x + 5}} = -11}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{{x^2}-x-5 = 0,\,\,\,\,\,\,\,\,\,\,}\\{{x^2} + 11x + 55 = 0,}\end{array}} \right.}\\{x \ne -5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{{1 \pm \sqrt {21} }}{2}.\)
Ответ: \(\dfrac{{1 \pm \sqrt {21} }}{2}.\)