6) \(\frac{7}{x-3} + 1 = \frac{18}{x^2-6x+9}\)
\(\frac{7}{x-3} + 1 = \frac{18}{(x-3)^2}\)
Приведем к общему знаменателю \((x-3)^2\):
\(\frac{7(x-3)}{(x-3)^2} + \frac{(x-3)^2}{(x-3)^2} = \frac{18}{(x-3)^2}\)
\(7(x-3) + (x-3)^2 = 18\)
\(7x - 21 + x^2 - 6x + 9 = 18\)
\(x^2 + x - 12 - 18 = 0\)
\(x^2 + x - 30 = 0\)
\(D = 1 - 4*1*(-30) = 1 + 120 = 121\)
\(x_1 = \frac{-1 + 11}{2} = 5\)
\(x_2 = \frac{-1 - 11}{2} = -6\)
Ответ: \(x_1 = 5, x_2 = -6\)