Решите уравнение, обозначив одну из взаимно обратных дробей через t, а другую – через 1/t: (x^2-4)/x+x/(x^2-4)=3*1/3.

\[\frac{x^{2} - 4}{x} + \frac{x}{x^{2} - 4} = 3\frac{1}{3}\]

\[t = \frac{x^{2} - 4}{x};\ \ \ \frac{1}{t} = \frac{x}{x^{2} - 4}:\]

\[t + \frac{1}{t} = \frac{10}{3}\ \ \ \ \ \ \ \ \ | \cdot 3t\]

\[3t^{2} + 3 - 10t = 0\]

\[3t^{2} - 10t + 3 = 0\]

\[D = 100 - 36 = 64\]

\[t_{1} = \frac{10 + 8}{6} = 3;\ \ \ \ \]

\[t_{2} = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3}.\]

\[Подставим:\]

\[1)\ \frac{x^{2} - 4}{x} = 3\ \ \ \ \ \ \ \ | \cdot x\]

\[x^{2} - 4 = 3x\]

\[x^{2} - 3x - 4 = 0\]

\[x_{1} + x_{2} = 3;\ \ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = 4;\ \ \ x_{2} = - 1.\]

\[2)\ \frac{x^{2} - 4}{x} = \frac{1}{3}\ \ \ \ \ \ | \cdot 3x\]

\[3x^{2} - 12 - x = 0\]

\[3x^{2} - x - 12 = 0\]

\[D = 1 + 144 = 145\]

\[x_{1,2} = \frac{1 \pm \sqrt{145}}{6}.\]

\[Ответ:x = - 1;\ \ x = 4;\ \ \]

\[x = \frac{1 \pm \sqrt{145}}{6}.\]