ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 949

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

\[1)\ \log_{3}(y - 3) - \log_{3}x = 0;\]

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

\[y - 3 = x\]

\[y = x + 3.\]

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

\[x^{2} + 2ax + a^{2} - 2x - 6 - 5a = 0\]

\[x^{2} + (2a - 2)x + \left( a^{2} - 5a - 6 \right) = 0\]

\[D = (2a - 2)^{2} - 4\left( a^{2} - 5a - 6 \right) \geq 0\]

\[4a^{2} - 8a + 4 - 4a^{2} + 20a + 24 \geq 0\]

\[12a + 28 \geq 0\]

\[12a \geq - 28\]

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

\[3)\ Область\ определения:\]

\[y > 3;\ \ \ x > 0.\]

\[4)\ Корень\ второго\ уравнения:\]

\[x = \frac{- (2a - 2) + \sqrt{12a + 28}}{2} > 0\]

\[\sqrt{12a + 28} > 2a - 2\]

\[12a + 28 > 4a^{2} - 8a + 4\]

\[4a^{2} - 20a - 24 < 0\]

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

\[D = 25 + 24 = 49\]

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

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

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

\[- 1 < a < 6.\]

\[2a - 2 < 0\ \ \ \]

\[a < 1.\]

\[Ответ:\ - \frac{7}{3} \leq a < 6.\]