Решите уравнение: x^2+3x=8/(x^2+3x-2) (подстановка y=x^2+3x).

Ответ:

\[x² + 3x = \frac{8}{x^{2} + 3x - 2}\]

\[ОДЗ:\ \ \]

\[x^{2} + 3x - 2 \neq 0\]

\[D = b^{2} - 4ac =\]

\[= 9 - 4 \cdot 1 \cdot ( - 2) = 9 + 8 = 17\]

\[x_{1} = \frac{- 3 + \sqrt{17}}{2};\ \ \ \ \ \ \]

\[x_{2} = \frac{- 3 - \sqrt{17}}{2}\]

\[Пусть\ y = x^{2} + 3x:\]

\[y = \frac{8}{y - 2}\]

\[y(y - 2) = 8\]

\[y^{2} - 2y - 8 = 0\]

\[D = b^{2} - 4ac =\]

\[= 4 - 4 \cdot 1 \cdot ( - 8) = 36\]

\[y_{1} = \frac{2 + 6}{2} = \frac{8}{2} = 4\]

\[y_{2} = \frac{2 - 6}{2} = - \frac{4}{2} = - 2\]

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

\[x^{2} + 3x - 4 = 0\]

\[D = b^{2} - ac =\]

\[= 9 - 4 \cdot 1 \cdot ( - 4) = 25\]

\[x_{1} = \frac{- 3 + 5}{2} = \frac{2}{2} = 1\]

\[x_{2} = \frac{- 3 - 5}{2} = - \frac{8}{2} = - 4\]

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

\[x^{2} + 3x + 2 = 0\]

\[D = b^{2} - 4ac = 9 - 4 \cdot 1 \cdot 2 = 1\]

\[x_{1} = \frac{- 3 + 1}{2} = - \frac{2}{2} = - 1\]

\[x_{2} = \frac{- 3 - 1}{2} = - \frac{4}{2} = - 2\]

\[Ответ:x = 1;\ \ x = - 4;\ \ \]

\[x = - 1;\ \ x = - 2.\]