Найдите корни уравнения 16/(x^2+x)-6/(x^2-x)=1/x.

\[\frac{16}{x^{2} + x} - \frac{6}{x^{2} - x} = \frac{1}{x}\]

\[\frac{16^{\backslash x - 1}}{x(x + 1)} - \frac{6^{\backslash x + 1}}{x(x - 1)} - \frac{1^{\backslash x^{2} - 1}}{x} = 0\]

\[\frac{16x - 16 - 6x - 6 - x^{2} + 1}{x(x + 1)(x - 1)} = 0\]

\[ОДЗ:\ \ x \neq 0;\ \ \ x \neq \pm 1.\]

\[- x^{2} + 10x - 21 = 0\ \ \ \ \ |\ :( - 1)\]

\[x^{2} - 10x + 21 = 0\]

\[x_{1} + x_{2} = 10;\ \ \ \ x_{1} \cdot x_{2} = 21\]

\[x_{1} = 7;\ \ x_{2} = 3.\]

\[Ответ:\ \ x = 3;\ \ x = 7.\]