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