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