Решите уравнение: 1/(x+6)+3/(x^2-6x)=72/(x^3-36x).

\[\frac{1}{x + 6} + \frac{3}{x^{2} - 6x} = \frac{72}{x^{3} - 36x}\]

\[\frac{1}{x + 6} + \frac{3}{x(x - 6)} = \frac{72}{x\left( x^{2} - 36 \right)}\]

\[x^{2} - 6x + 3x + 18 - 72 = 0\]

\[x² - 3x - 54 = 0\]

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

\[x_{1} \cdot x_{2} = - 54\]

\[\Longrightarrow x_{1} = 9;\ \ \]

\[x_{2} = - 6\ (не\ подходит).\]

\[Ответ:x = 9.\]