Решите уравнение методом замены переменной: 1/(x^2-3x+3)+2/(x^2-3x+4)=6/(x^2-3x+5).

\[\frac{1}{x^{2} - 3x + 3} + \frac{2}{x^{2} - 3x + 4} =\]

\[= \frac{6}{x^{2} - 3x + 5}\ \]

\[Пусть\ t = x^{2} - 3x + 3:\]

\[- 3t^{2} + t + 2 = 0\]

\[3t^{2} - t - 2 = 0\]

\[t_{1} + t_{2} = \frac{1}{3}\]

\[t_{1} \cdot t_{2} = - \frac{2}{3}\]

\[\Longrightarrow t_{1} = 1,\ \ t_{2} = - \frac{2}{3}.\]

\[Ответ:x = 2;x = 1.\]