\[x² + 3x = \frac{8}{x^{2} + 3x - 2}\]
\[ОДЗ:\ \ \]
\[x^{2} + 3x - 2 \neq 0\]
\[D = b^{2} - 4ac =\]
\[= 9 - 4 \cdot 1 \cdot ( - 2) = 9 + 8 = 17\]
\[x_{1} = \frac{- 3 + \sqrt{17}}{2};\ \ \ \ \ \ \]
\[x_{2} = \frac{- 3 - \sqrt{17}}{2}\]
\[Пусть\ y = x^{2} + 3x:\]
\[y = \frac{8}{y - 2}\]
\[y(y - 2) = 8\]
\[y^{2} - 2y - 8 = 0\]
\[D = b^{2} - 4ac =\]
\[= 4 - 4 \cdot 1 \cdot ( - 8) = 36\]
\[y_{1} = \frac{2 + 6}{2} = \frac{8}{2} = 4\]
\[y_{2} = \frac{2 - 6}{2} = - \frac{4}{2} = - 2\]
\[x^{2} + 3x = 4\]
\[x^{2} + 3x - 4 = 0\]
\[D = b^{2} - ac =\]
\[= 9 - 4 \cdot 1 \cdot ( - 4) = 25\]
\[x_{1} = \frac{- 3 + 5}{2} = \frac{2}{2} = 1\]
\[x_{2} = \frac{- 3 - 5}{2} = - \frac{8}{2} = - 4\]
\[x^{2} + 3x = - 2\]
\[x^{2} + 3x + 2 = 0\]
\[D = b^{2} - 4ac = 9 - 4 \cdot 1 \cdot 2 = 1\]
\[x_{1} = \frac{- 3 + 1}{2} = - \frac{2}{2} = - 1\]
\[x_{2} = \frac{- 3 - 1}{2} = - \frac{4}{2} = - 2\]
\[Ответ:x = 1;\ \ x = - 4;\ \ \]
\[x = - 1;\ \ x = - 2.\]