Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 632

\[\boxed{\mathbf{632}.}\]

\[1)\ \sqrt{x^{2} - 3x + 2} > x + 3;\]

\[x^{2} - 3x + 2 > x^{2} + 6x + 9;\]

\[- 9x > 7;\]

\[x < - \frac{7}{9};\]

\[Выражение\ имеет\ смысл\ при:\]

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

\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]

\[x_{1} = \frac{3 - 1}{2} = 1\ \ и\ \]

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

\[(x - 1)(x - 2) \geq 0;\]

\[x \leq 1\ \ и\ \ x \geq 2;\]

\[Неравенство\ всегда\ верно\ при:\]

\[x - 3 \leq 0;\]

\[x \leq 3;\]

\[Ответ:\ \ x < - \frac{7}{9}.\]

\[2)\ \sqrt{2x^{2} - 7x - 4} > - x - \frac{1}{4};\]

\[\sqrt{2x^{2} - 7x - 4} > - \left( x + \frac{1}{4} \right);\]

\[2x^{2} - 7x - 4 > x^{2} + \frac{1}{2}x + \frac{1}{16}\text{\ \ \ \ \ }\]

\[| \bullet 16;\]

\[32x^{2} - 112x - 64 > 16x^{2} +\]

\[+ 8x + 1;\]

\[16x^{2} - 120x - 65 > 0;\]

\[D = 120^{2} + 4 \bullet 16 \bullet 65 =\]

\[= 14\ 400 + 4\ 160 =\]

\[= 18\ 560 = 64 \bullet 290\]

\[x = \frac{120 \pm \sqrt{18\ 560}}{2 \bullet 16} =\]

\[= \frac{120 \pm 8\sqrt{290}}{32} = \frac{15 \pm \sqrt{290}}{4};\]

\[\left( x - \frac{15 - \sqrt{290}}{4} \right)\left( x - \frac{15 + \sqrt{290}}{4} \right) > 0;\]

\[x < \frac{15 - \sqrt{290}}{4}\text{\ \ }и\ \]

\[\ x > \frac{15 + \sqrt{290}}{4};\]

\[Выражение\ имеет\ смысл\ при:\]

\[2x^{2} - 7x - 4 \geq 0;\]

\[D = 7^{2} + 4 \bullet 2 \bullet 4 =\]

\[= 49 + 32 = 81\]

\[x_{1} = \frac{7 - 9}{2 \bullet 2} = - \frac{2}{4} = - 0,5;\]

\[x_{2} = \frac{7 + 9}{2 \bullet 2} = \frac{16}{4} = 4;\]

\[(x + 0,5)(x - 4) \geq 0;\]

\[x \leq - 0,5\ \ и\ \ x \geq 4;\]

\[Неравенство\ всегда\ верно\ при:\]

\[- x - \frac{1}{4} \leq 0;\]

\[x + \frac{1}{4} \geq 0;\]

\[x \geq - \frac{1}{4};\]

\[Ответ:\ \ x < \frac{15 - \sqrt{290}}{4};\ \ x \geq 4.\]