ГДЗ по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 34

\[\boxed{\mathbf{34.}}\]

\[\frac{x}{x - a} + \frac{1}{x + a} = \frac{2}{x^{2} - a^{2}}\]

\[\frac{x^{\backslash\text{(}x + a)}}{x - a} + \frac{1^{\backslash(x - a)}}{x + a} - \frac{2}{(x - a)(x + a)} = 0\]

\[\frac{x^{2} + ax + x - a - 2}{(x - a)(x + a)} = 0\]

\[\frac{x^{2} + (a + 1)x - (a + 2)}{x^{2} - a^{2}} = 0\]

\[\left\{ \begin{matrix} x^{2} + (a + 1)x - (a + 2) = 0 \\ x^{2} - a^{2} \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} + (a + 1)x - (a + 2) = 0\]

\[Уравнение\ имеет\ один\ корень\]

\[\ при\ D = 0.\]

\[x^{2} + (a + 1)x - (a + 2) = 0\]

\[D = (a + 1)^{2} + 4 \cdot (a + 2) =\]

\[= a^{2} + 2a + 1 + 4a + 8 =\]

\[= a^{2} + 6a + 9 =\]

\[= (a + 3)^{2}\]

\[(a + 3)^{2} = 0\]

\[a + 3 = 0\]

\[a = - 3.\]

\[При\ a = - 3:\]

\[x = \frac{- (a + 1)}{2}.\]

\[При\ a \neq - 3:\]

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

\[= \frac{- 2a - 4}{2} = - a - 2;\]

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

\[x^{2} - a^{2} \neq 0\]

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

\[x \neq a;\ \ a \neq - a.\]

\[При\ x = - a - 2 = a:\]

\[a = - 1.\]

\[При\ x = - a - 2 = - a:\]

\[решений\ нет.\]

\[Ответ:при\ a = - 3;a = - 1.\]