Решите уравнение: 1/(x-3)^2-3/(x-3)-4=0.

\[\frac{1}{(x - 3)^{2}} - \frac{3}{x - 3} - 4 = 0\ \ | \cdot (x - 3)^{2}\]

\[1 - 3(x - 3) - 4(x - 3)^{2} = 0\]

\[1 - 3x + 3 - 4\left( x^{2} - 6x + 9 \right) = 0\]

\[4 - 3x - 4x^{2} + 24x - 36 = 0\]

\[- 4x^{2} + 21x - 32 = 0\]

\[4x^{2} - 21x + 32 = 0\]

\[D = 441 - 512 < 0\]

\[Ответ:нет\ корней.\]