\(3\,\left( {x + \dfrac{1}{{{x^2}}}} \right)-7\,\left( {1 + \dfrac{1}{x}} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3x\left( {1 + \dfrac{1}{{{x^3}}}} \right)-7\left( {x + \dfrac{1}{x}} \right) = 0.\)
Воспользуемся формулой:
\({a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2}-ab + {b^2}} \right).\)
\(3x\left( {1 + \dfrac{1}{x}} \right)\left( {1-\dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right)-7\left( {x + \dfrac{1}{x}} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left( {1 + \dfrac{1}{x}} \right)\left( {3x-3 + \dfrac{3}{x}-7} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{\left( {x + 1} \right)\left( {3{x^2}-10x + 3} \right)}}{{{x^2}}} = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{x + 1 = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{3{x^2}-10x + 3 = 0,}\end{array}} \right.}\\{x \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = \dfrac{1}{3},\,}\\{x = 3,\,\,}\end{array}} \right.}\\{x \ne 0\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = \dfrac{1}{3},}\\{x = 3.\,\,}\end{array}} \right.\)
Ответ: \(-1;\;\;\dfrac{1}{3};\;\;3.\)