Решите уравнение: (x-2)(x+7)=19/(x+1)(x+4) (подстановка y=x^2+5x).

\[(x - 2)(x + 7) = \frac{19}{(x + 1)(x + 4)}\text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[ОДЗ:x
eq - 1;\ \ x
eq - 4\]

\[x^{2} + 5x - 14 = \frac{19}{x^{2} + 5x + 4}\]

\[Пусть\ y = x^{2} + 5x:\]

\[y - 14 = \frac{19}{y + 4}\]

\[(y - 14)(y + 4) = 19\]

\[y^{2} - 14y + 4y - 56 = 19\]

\[y^{2} - 10y - 56 - 19 = 0\]

\[y^{2} - 10y - 75 = 0\]

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

\[= 100 - 4 \cdot 1 \cdot ( - 75) = 400\]

\[y_{1} = \frac{10 + 20}{2} = \frac{30}{2} = 15\]

\[y_{2} = \frac{10 - 20}{2} = - \frac{10}{2} = - 5\]

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

\[x^{2} + 5x + 5 = 0\]

\[D = 25 - 20 = 5\]

\[x_{1,2} = \frac{- 5 \pm \sqrt{5}}{2}.\]

\[x^{2} + 5x - 15 = 0\]

\[D = b^{2} - 4ac = 25 + 60 = 85.\]

\[x_{3,4} = \frac{- 5 \pm \sqrt{85}}{2}\]

\[Ответ:\ \ x = \frac{- 5 \pm \sqrt{5}}{2};\ \ \ \ \]

\[x = \frac{- 5 \pm \sqrt{85}}{2}.\]