Задача 4. Решите уравнение \(\frac{1}{{{{\left( {x-3} \right)}^2}}} = \frac{3}{{x-3}} + 4\)
ОТВЕТ: 2; 13/4.
\(\frac{1}{{{{\left( {x-3} \right)}^2}}} = \frac{3}{{x-3}} + 4\) Пусть \(\frac{1}{{x-3}} = t\). Тогда уравнение примет вид: \({t^2} = 3t + 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,{t^2}-3t-4 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -1,\,\,\,}\\{t = 4.\,\,\,\,}\end{array}} \right.\) Вернёмся к прежней переменной: \(\left[ {\begin{array}{*{20}{c}}{\frac{1}{{x-3}} = -1}\\{\frac{1}{{x-3}} = 4\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{x-3 = -1,}\\{x-3 = \frac{1}{4},}\end{array}} \right.}\\{x \ne 3\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{x = 2,\,\,\,\,}\\{x = \frac{{13}}{4},}\end{array}} \right.}\\{x \ne 3\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 2,\,\,\,\,}\\{x = \frac{{13}}{4}.}\end{array}} \right.\) Ответ: \(2;\;\;\frac{{13}}{4}.\)