Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 383

\[\boxed{\mathbf{383}\mathbf{.}}\]

\[1)\lg\left( x^{2} + 2x + 2 \right) < 1\]

\[\lg\left( x^{2} + 2x + 2 \right) < \lg 10\]

\[x^{2} + 2x + 2 < 10\]

\[x^{2} + 2x - 8 < 0\]

\[D = 2^{2} + 4 \bullet 8 = 4 + 32 = 36\]

\[x_{1} = \frac{- 2 - 6}{2} = - 4;\text{\ \ }\]

\[x_{2} = \frac{- 2 + 6}{2} = 2.\]

\[(x + 4)(x - 2) < 0\]

\[- 4 < x < 2.\]

\[имеет\ смысл\ при:\]

\[x^{2} + 2x + 2 > 0\]

\[D = 2^{2} - 4 \bullet 2 = 4 - 8 = - 4 < 0\]

\[a = 1 > 0 \Longrightarrow x - любое\ число.\]

\[Ответ:\ \ - 4 < x < 2.\]

\[2)\log_{3}\left( x^{2} + 7x - 5 \right) > 1\]

\[\log_{3}\left( x^{2} + 7x - 5 \right) > \log_{3}3\]

\[x^{2} + 7x - 5 > 3\]

\[x^{2} + 7x - 8 > 0\]

\[D = 7^{2} + 4 \bullet 8 = 49 + 32 = 81\]

\[x_{1} = \frac{- 7 - 9}{2} = - 8;\text{\ \ }\]

\[x_{2} = \frac{- 7 + 9}{2} = 1.\]

\[(x + 8)(x - 1) > 0\]

\[x < - 8;\text{\ \ }x > 1.\]

\[имеет\ смысл\ при:\]

\[x^{2} + 7x - 5 > 0\]

\[D = 7^{2} + 4 \bullet 5 = 49 + 20 = 69\]

\[x_{1} = \frac{- 7 - \sqrt{69}}{2} \approx \frac{- 7 - 8,3}{2} \approx\]

\[\approx - 7,6;\]

\[x_{2} = \frac{- 7 + \sqrt{69}}{2} \approx \frac{- 7 + 8,3}{2} \approx\]

\[\approx 0,6.\]

\[(x + 7,6)(x - 0,6) > 0\]

\[x < - 7,6;\text{\ \ }x > 0,6.\]

\[Ответ:\ \ x < - 8;\ \ x > 1.\]