1 способ
Запишем область допустимых значений: \(x \ne 0.\)
\(\dfrac{1}{{{x^2}}} + \dfrac{2}{x}-3 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{1 + 2x-3{x^2}}}{{{x^2}}} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}1 + 2x-3{x^2} = 0,\\x \ne 0.\end{array} \right.\)
\(1 + 2x-3{x^2} = 0\left| { \cdot \left( {-1} \right)} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,3{x^2}-2x-1 = 0;\,\,\,\,\,\,\,D = {\left( {-2} \right)^2}-4 \cdot 3 \cdot \left( {-1} \right) = 16.\)
\(\left\{ \begin{array}{l}\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{2 + 4}}{{2 \cdot 3}} = 1,\,\,\,\,}\\{x = \dfrac{{2-4}}{{2 \cdot 3}} = -\dfrac{1}{3},}\end{array}} \right.\\x \ne 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 1,\\x = -\dfrac{1}{3}.\end{array} \right.\)
Ответ: \(-\dfrac{1}{3};\,\,\,\,1.\)
2 способ
Запишем область допустимых значений: \(x \ne 0.\)
Пусть \(\dfrac{1}{x} = t,\) тогда уравнение примет вид:
\({t^2} + 2t-3 = 0;\,\,\,\,\,\,D = {2^2}-4 \cdot 1 \cdot \left( {-3} \right) = 16;\,\,\,\,\,\,\,\left[ \begin{array}{l}t = \dfrac{{-2 + 4}}{2} = 1,\\t = \dfrac{{-2-4}}{2} = -3.\end{array} \right.\)
Возвращаясь к прежней переменной, получим:
\(\left[ \begin{array}{l}\dfrac{1}{x} = 1,\\\dfrac{1}{x} = -3\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 1,\\x = -\dfrac{1}{3}.\end{array} \right.\)
Ответ: \(-\dfrac{1}{3};\,\,\,\,1.\)