Задача 1. Решите уравнение \(\dfrac{1}{{{x^2}}} + \dfrac{2}{x}-3 = 0\)
ОТВЕТ: -1/3; 1.
\(\dfrac{1}{{{x^2}}} + \dfrac{2}{x}-3 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{1 + 2x-3{x^2}}}{{{x^2}}} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{1 + 2x-3{x^2} = 0}\\{{x^2} \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,\,\,}\\{x = -\dfrac{1}{3},}\end{array}} \right.}\\{x \ne 0\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,\,\,}\\{x = -\dfrac{1}{3}.}\end{array}} \right.\) Ответ: \(-\dfrac{1}{3};\;\;1.\)