Найдите корни уравнения: 7/(x-3)+1=18/(x^2-6x+9).

Ответ:

\[\frac{7}{x - 3} + 1 = \frac{18}{x² - 6x + 9}\]

\[ОДЗ:\ \ x - 3 \neq 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \Longrightarrow x \neq 3\]

\[\frac{7}{x - 3} - \frac{18}{(x - 3)^{2}} = - 1\]

\[\frac{7 \cdot (x - 3) - 18}{(x - 3)²} = - 1\]

\[7x - 21 - 18 = - x^{2} + 6x - 9\]

\[x^{2} + 7x - 6x - 39 + 9 = 0\]

\[x^{2} + x - 30 = 0\]

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

\[x_{1} \cdot x_{2} = - 30 \Longrightarrow x_{1} = - 6;\ \ \]

\[x_{2} = 5\]

\[Ответ:x = - 6\ \ и\ \ x = 5.\]