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

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

\[Пусть\ y = \frac{x^{2} - 3}{x}:\]

\[y + \frac{1}{y} = \frac{5}{2}\]

\[y^{2} + 1 - 2,5y = 0\ \ \ \ \ | \cdot 2\]

\[2y^{2} - 5y + 2 = 0\]

\[D = 25 - 16 = 9\]

\[y_{1} = \frac{5 + 3}{4} = 2;\ \ \ \]

\[y_{2} = \frac{5 - 3}{4} = 0,5.\]

\[Подставим:\]

\[1)\ \frac{x^{2} - 3}{x} = 2^{\backslash x}\]

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

\[D = 1 + 3 = 4\]

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

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

\[2)\ \frac{x^{2} - 3}{x} = {0,5}^{\backslash x}\]

\[x^{2} - 0,5x - 3 = 0\ \ \ \ \ | \cdot 2\]

\[2x^{2} - x - 6 = 0\]

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{1 + 7}{4} = 2;\ \ \]

\[\ x_{2} = \frac{1 - 7}{4} = - 1,5\]

\[Ответ:x = - 1,5;\ \ x = - 1;\ \]

\[\ x = 2;\ \ x = 3.\]