1 способ
Запишем область допустимых значений: \(x \ne 0.\)
\(\dfrac{1}{{{x^2}}} + \dfrac{4}{x}-12 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{1 + 4x-12{x^2}}}{{{x^2}}} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}1 + 4x-12{x^2} = 0,\\x \ne 0.\end{array} \right.\)
\(1 + 4x-12{x^2} = 0\left| { \cdot \left( {-1} \right)} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,12{x^2}-4x-1 = 0;\,\,\,\,\,\,\,D = {\left( {-4} \right)^2}-4 \cdot 12 \cdot \left( {-1} \right) = 64.\)
\(\left\{ \begin{array}{l}\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{4 + 8}}{{2 \cdot 12}} = \dfrac{1}{2},\,\,\,\,}\\{x = \dfrac{{4-8}}{{2 \cdot 12}} = -\dfrac{1}{6},}\end{array}} \right.\\x \ne 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{1}{2},\\x = -\dfrac{1}{6}.\end{array} \right.\)
Ответ: \(-\dfrac{1}{6};\,\,\,\,\dfrac{1}{2}.\)
2 способ
Запишем область допустимых значений: \(x \ne 0.\)
Пусть \(\dfrac{1}{x} = t,\) тогда уравнение примет вид:
\({t^2} + 4t-12 = 0;\,\,\,\,\,\,D = {4^2}-4 \cdot 1 \cdot \left( {-12} \right) = 64;\,\,\,\,\,\,\,\,\left[ \begin{array}{l}t = \dfrac{{-4 + 8}}{2} = 2,\\t = \dfrac{{-4-8}}{2} = -6.\end{array} \right.\)
Возвращаясь к прежней переменной, получим:
\(\left[ \begin{array}{l}\dfrac{1}{x} = 2,\\\dfrac{1}{x} = -6\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{1}{2},\\x = -\dfrac{1}{6}.\end{array} \right.\)
Ответ: \(-\dfrac{1}{6};\,\,\,\,\dfrac{1}{2}.\)