Запишем область допустимых значений: \(x-3 \ne 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \ne 3.\)
Пусть \(\dfrac{1}{{x-3}} = t,\) тогда уравнение примет вид:
\({t^2}-3t-4 = 0;\,\,\,\,\,\,D = {\left( {-3} \right)^2}-4 \cdot 1 \cdot \left( {-4} \right) = 25;\,\,\,\,\,\,\,\left[ \begin{array}{l}t = \dfrac{{3 + 5}}{2} = 4,\\t = \dfrac{{3-5}}{2} = -1.\end{array} \right.\)
Возвращаясь к прежней переменной, получим:
\(\left[ \begin{array}{l}\dfrac{1}{{x-3}} = 4,\\\dfrac{1}{{x-3}} = -1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x-3 = \dfrac{1}{4},\\x-3 = -1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{13}}{4},\\x = 2.\end{array} \right.\)
Ответ: \(\dfrac{{13}}{4};\,\,\,\,2.\)