Решите уравнение: x^2+x+1=15/(x^2+x+3) (подстановка y=x^2+x+1).

\[x² + x + 1 = \frac{15}{x^{2} + x + 3}\]

\[Пусть\ y = x^{2} + x + 1:\]

\[y = \frac{15}{y + 2}\]

\[ОДЗ:\ \ y \neq - 2\]

\[y(y + 2) = 15\]

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

\[y_{1} + y_{2} = - 2\]

\[y_{1} \cdot y_{2} = - 15 \Longrightarrow y_{1} = - 5;\ \ y_{2} =\]

\[= 3\]

\[1)\ x² + x + 1 = - 5\]

\[x^{2} + x + 6 = 0\]

\[x_{1} \cdot x_{2} = - 1\]

\[x_{1} \cdot x_{2} = 6 \Longrightarrow нет\ решения.\]

\[D = b² - 4ac = 1 - 4 \cdot 6 < 0.\]

\[2)\ x² + x + 1 = 3\]

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

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

\[x_{1} \cdot x_{2} = - 2 \Longrightarrow x_{1} = - 2;\ \ x_{2} =\]

\[= 1.\]

\[Ответ:x = - 2;\ \ x = 1.\]