Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1330

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\[1)\ \frac{3x - 1}{x + 2} - \frac{7}{2 + x} = \frac{7x^{2} - 28}{x^{2} - 4} + \frac{18}{2 - x}\]

\[4x^{2} - 4x - 80 = 0\]

\[x^{2} - x - 20 = 0\]

\[D = 1 + 80 = 81\]

\[x_{1} = \frac{1 - 9}{2} = - \frac{8}{2} = - 4;\]

\[x_{2} = \frac{1 + 9}{2} = \frac{10}{2} = 5.\]

\[Ответ:\ - 4;\ 5.\]

\[2)\ \frac{x + 1}{x + 3} - \frac{12}{x^{2} - 9} = \frac{2x - 1}{3 - x}\]

\[x^{2} - 2x - 15 = - 2x^{2} - 5x + 3\]

\[3x^{2} + 3x - 18 = 0\]

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

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{- 1 - 5}{2} = - \frac{6}{2} = - 3;\]

\[x_{2} = \frac{- 1 + 5}{2} = \frac{4}{2} = 2.\]

\[ОДЗ:\]

\[(x - 3)(x + 3) \neq 0\]

\[x_{1} \neq 3\ \ и\ \ x_{2} \neq - 3.\]

\[Ответ:\ \ x = 2.\]