Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 622

\[\boxed{\mathbf{622}.}\]

\[1)\ \sqrt{x + 1} \cdot \sqrt{x - 2} = a\ \]

\[при\ a < 0 \Longrightarrow нет\ корней;\]

\[при\ a = 0 \Longrightarrow x = 2.\]

\[\left( \sqrt{x + 1} \cdot \sqrt{x - 2} \right)^{2} = a^{2}\]

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

\[x^{2} - x - 2 = a^{2};\ \ \ a > 0:\]

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

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

\[= 1 + 8 + 4a^{2} = 4a^{2} + 9\]

\[4a^{2} + 9 > 0\]

\[x_{1} = \frac{1 + \sqrt{4a^{2} + 9}}{2};\ \ \]

\[x_{2} = \frac{1 - \sqrt{4a^{2} - 9}}{2}.\]

\[ОДЗ:\ \ x + 1 \geq 0;\ \ \]

\[\ \ x - 2 \geq 0 \Longrightarrow x \geq 2.\]

\[При\ a < 0 - нет\ решений;\]

\[при\ a = 0:\ \ x_{1} = \frac{1 + \sqrt{9}}{2} = 2;\ \]

\[\text{\ \ }x_{2} = \frac{1 - \sqrt{9}}{2} =\]

\[= - 1\ (не\ подходит);\]

\[при\ a > 0:\]

\[\left\{ \begin{matrix} \frac{1 + \sqrt{4a^{2} + 9}}{2} \geq 2 \\ \frac{1 - \sqrt{4a^{2} + 9}}{2} \geq 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 1 + \sqrt{4a^{2} + 9} \geq 4 \\ 1 - \sqrt{4a^{2} + 9} \geq 4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \sqrt{4a^{2} + 9} \geq 3\ \ \ \ (1) \\ \sqrt{4a^{2} + 9} \leq - 3\ (2) \\ \end{matrix} \right.\ \]

\[(1)\ 4a^{2} + 9 \geq 9\]

\[4a^{2} \geq 9\]

\[a^{2} \geq 0.\]

\[(2)\ 4a^{2} + 9 \leq - 9\]

\[4a^{2} \leq - 18\]

\[корней\ нет.\]

\[Ответ:\]

\[при\ a < 0 - решений\ нет;\]

\[при\ a = 0:\ \ x = 2;\]

\[при\ a > 0:\ \ x = \frac{1 + \sqrt{9 + 4a^{2}}}{2}.\]

\[2)\ \sqrt{x} \cdot \sqrt{x + 2} = a - 1\]

\[При\ a - 1 < 0;\ \ \]

\[a < 1 - решений\ нет.\]

\[При\ a = 1:\]

\[\sqrt{x} \cdot \sqrt{x + 2} = 0\]

\[x(x + 2) = 0\]

\[x = 0;\ \ \ x = - 2\ (не\ подходит).\]

\[При\ a > 1:\]

\[\left( \sqrt{x(x + 2} \right)^{2} = (a - 1)^{2}\]

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

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

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

\[x_{1} = - 1 + \sqrt{a^{2} - 2a + 2};\ \]

\[\ x_{2} = - 1 - \sqrt{a^{2} - 2a + 2}.\]

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

\[(1)\ - 1 + \sqrt{a^{2} - 2a + 2} \geq 0\]

\[\left( \sqrt{a^{2} - 2a + 2} \right)^{2} \geq 1\]

\[a^{2} - 2a + 2 - 1 \geq 0\]

\[(a - 1)^{2} \geq 0\]

\[a - любое\ число.\]

\[(2)\ - 1 - \sqrt{a^{2} - 2a + 2} \geq 0\]

\[- \left( \sqrt{a^{2} - 2a + 2} \right)^{2} \geq 1\]

\[a^{2} - 2a + 2 \leq - 1\]

\[нет\ корней.\]

\[Ответ:\]

\[при\ a < 1 - решений\ нет;\]

\[при\ a = 1:\ \ x = 0;\]

\[при\ a > 1:\ \ \]

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