\[\frac{x^{\backslash x}}{3 - x} + 1^{x(3 - x)} = \frac{6^{\backslash 3 - x}}{x}\]
\[\frac{x^{2} + 3x - x^{2} - 18 + 6x}{x(3 - x)} = 0\]
\[\frac{9x - 18}{x(3 - x)} = 0;\ \ x \neq 0;\ \ x \neq 3\]
\[9x - 18 = 0\]
\[9x = 18\]
\[x = 2.\]
\[Ответ:при\ x = 2.\]