ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 928

\[1)\ \frac{13 - 3x + \sqrt{x^{2} - x - 6}}{5 - x} > 1\]

\[x < 5:\]

\[13 - 3x + \sqrt{x^{2} - x - 6} > 5 - x\]

\[\sqrt{x^{2} - x - 6} > 2x - 8\]

\[x^{2} - x - 6 > 4x^{2} - 32x + 64\]

\[3x^{2} - 31x + 70 < 0\]

\[D = 961 - 840 = 121\]

\[x_{1} = \frac{31 - 11}{2 \bullet 3} = \frac{10}{3};\]

\[x_{2} = \frac{31 + 11}{2 \bullet 3} = 7;\]

\[\left( x - \frac{10}{3} \right)(x - 7) < 0\]

\[\frac{10}{3} < x < 7\]

\[x < 7.\]

\[2x - 8 < 0\]

\[x < 4.\]

\[x > 5:\]

\[13 - 3x + \sqrt{x^{2} - x - 6} < 5 - x\]

\[x < \frac{10}{3}\]

\[x > 7;\ \ x > 4.\]

\[Область\ определения:\]

\[x^{2} - x - 6 \geq 0\]

\[D = 1 + 24 = 25\]

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

\[x_{2} = \frac{1 + 5}{2} = 3;\]

\[(x + 2)(x - 3) \geq 0\]

\[x \leq - 2;\ \ \ x \geq 3.\]

\[Ответ:\ \ \]

\[x \in ( - \infty;\ - 2\rbrack \cup \lbrack 3;\ 5) \cup (7;\ + \infty).\]

\[2)\ \frac{7 - 3x + \sqrt{x^{2} + 3x - 4}}{x - 3} < - 1\]

\[x > 3:\]

\[7 - 3x + \sqrt{x^{2} + 3x - 4} < 3 - x\]

\[\sqrt{x^{2} + 3x - 4} < 2x - 4\]

\[x^{2} + 3x - 4 < 4x^{2} - 16x + 16\]

\[3x^{2} - 19x + 20 > 0\]

\[D = 361 - 240 = 121\]

\[x_{1} = \frac{19 - 11}{2 \bullet 3} = \frac{4}{3};\]

\[x_{2} = \frac{19 + 11}{2 \bullet 3} = 5;\]

\[\left( x - \frac{4}{3} \right)(x - 5) > 0,\ \ \ \]

\[x < \frac{4}{3};\]

\[x > 5.\]

\[2x - 4 > 0\]

\[x > 2.\]

\[x < 3:\]

\[7 - 3x + \sqrt{x^{2} + 3x - 4} > 3 - x\]

\[\left( x - \frac{4}{3} \right)(x - 5) < 0\]

\[\frac{4}{3} < x < 5\]

\[x < 5.\]

\[2x - 4 < 0\]

\[x < 2.\]

\[Область\ определения:\]

\[x^{2} + 3x - 4 \geq 0\]

\[D = 9 + 16 = 25\]

\[x_{1} = \frac{- 3 - 5}{2} = - 4;\]

\[x_{2} = \frac{- 3 + 5}{2} = 1;\]

\[(x + 4)(x - 1) \geq 0\]

\[x \leq - 4;\ \ \ x \geq 1.\]

\[Ответ:\ \ \]

\[x \in ( - \infty;\ - 4\rbrack \cup \lbrack 1;\ 3) \cup (5;\ + \infty).\]