Решите уравнение методом замены переменной: 2/(x^2+3x+4)+3/(x^2+3x+1)=8/(x^2+3x-2).

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

\[= \frac{8}{x^{2} + 3x - 2}\]

\[Пусть\ \ \ t = x^{2} + 3x - 2:\]

\[2t(t + 3) + 3t(t + 6) =\]

\[= 8 \cdot (t + 6)(t + 3)\]

\[2t^{2} + 6t + 3t^{2} + 18t =\]

\[= 8 \cdot (t^{2} + 9t + 18)\]

\[5t^{2} + 24t = 8t^{2} + 72t + 144\]

\[5t^{2} + 24t - 8t^{2} - 72t - 144 = 0\]

\[- 3t^{2} - 48t - 144 = 0\ \ \ |\ :( - 3)\]

\[t² + 16t + 48 = 0\]

\[t_{1} + t_{2} = - 16;\ \ t_{1} \cdot t_{2} = 48\]

\[t_{1} = - 12,\ \ t_{2} = - 4.\]