ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1609

\[\boxed{\mathbf{1609}\mathbf{.}}\]

\[\left\{ \begin{matrix} \log_{3}(y - 3) - 2\log_{9}x = 0 \\ (x + a)^{2} - 2y - 5a = 0\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\log_{3}(y - 3) - 2\log_{9}x = 0\]

\[\log_{3}(y - 3) - \log_{3}x = 0\]

\[y - 3 = x.\]

\[Подставим:\]

\[(y - 3 + a)^{2} - 2y - 5a = 0\]

\[y^{2} + 2ay - 8y + a^{2} - 11a + 9 = 0\]

\[y^{2} + (2a - 8)y + \left( a^{2} - 11a + 9 \right) = 0\]

\[D = (2a - 8)^{2} - 4\left( a^{2} - 11a + 9 \right) =\]

\[= 4a^{2} - 32a + 64 - 4a^{2} + 44a - 36 =\]

\[= 12a + 28 = 4(3a + 7).\]

\[При\ D \geq 0:\]

\[3a + 7 \geq 0\]

\[3a \geq - 7\]

\[a \geq - \frac{7}{3}.\]

\[y = \frac{- (2a - 8) \pm 2\sqrt{3a + 7}}{2} =\]

\[= 4 - a \pm \sqrt{3a + 7}.\text{\ \ }\]

\[1)\ y - 3 > 0\]

\[4 - a - \sqrt{3a + 7} - 3 > 0\]

\[1 - a > \sqrt{3a + 7}\]

\[1 - 2a + a^{2} > 3a + 7\]

\[a^{2} - 5a - 6 > 0\]

\[D = 25 + 24 = 49\]

\[a_{1} = \frac{5 - 7}{2} = - 1;\]

\[a_{2} = \frac{5 + 7}{2} = 6;\]

\[(a + 1)(a - 6) > 0\]

\[a < - 1\ \ и\ \ a > 6.\]

\[При\ (1 - a) \geq 0:\]

\[- \frac{7}{3} \leq a < - 1.\]

\[2)\ 4 - a + \sqrt{3a + 7} - 3 > 0\]

\[\sqrt{3a + 7} > a - 1\]

\[3a + 7 > a^{2} - 2a + 1\]

\[a^{2} - 5a - 6 < 0\]

\[(a + 1)(a - 6) < 0\]

\[- 1 < a < 6.\]

\[Ответ:\ \ - \frac{7}{3} \leq a < 6.\]