ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 629

\[\boxed{\mathbf{629}.}\]

\[1)\ \sqrt{x + 1} - \sqrt{x} < \sqrt{x - 1};\]

\[\sqrt{x + 1} - \sqrt{x - 1} < \sqrt{x};\]

\[x + 1 - 2\sqrt{(x + 1)(x - 1)} +\]

\[+ x - 1 < x;\]

\[x < 2\sqrt{x^{2} - 1};\]

\[x^{2} < 4\left( x^{2} - 1 \right);\]

\[x^{2} < 4x^{2} - 4;\]

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

\[x^{2} > \frac{4}{3};\]

\[x < - \frac{2}{\sqrt{3}}\text{\ \ }и\ \ x > \frac{2}{\sqrt{3}};\]

\[Выражение\ имеет\ смысл\ при:\]

\[x + 1 \geq 0\ \ \ \Longrightarrow \ \ \ x \geq - 1;\]

\[x - 1 \geq 0\ \ \ \Longrightarrow \ \ \ x \geq 1;\]

\[x \geq 0;\]

\[Ответ:\ \ x > \frac{2}{\sqrt{3}}.\]

\[2)\ \sqrt{x + 3} < \sqrt{7 - x} + \sqrt{10 - x};\]

\[x + 3 < 7 - x +\]

\[+ 2\sqrt{(7 - x)(10 - x)} + 10 - x;\]

\[3x -\]

\[- 14 < 2\sqrt{70 - 7x - 10x + x^{2}};\]

\[9x^{2} - 84x + 196 <\]

\[< 4\left( 70 - 17x + x^{2} \right);\]

\[9x^{2} - 84x + 196 < 280 -\]

\[- 68x + 4x^{2};\]

\[5x^{2} - 16x - 84 < 0;\]

\[D = 16^{2} + 4 \bullet 5 \bullet 84 =\]

\[= 256 + 1680 = 1936\]

\[x_{1} = \frac{16 - 44}{2 \bullet 5} = - \frac{28}{10} = - 2,8;\]

\[x_{2} = \frac{16 + 44}{2 \bullet 5} = \frac{60}{10} = 6;\]

\[(x + 2,8)(x - 6) < 0;\]

\[- 2,8 < x < 6;\]

\[Выражение\ имеет\ смысл\ при:\]

\[x + 3 \geq 0\ \ \ \Longrightarrow \ \ \ x \geq - 3;\]

\[7 - x \geq 0\ \ \ \Longrightarrow \ \ x \leq 7;\]

\[10 - x \geq 0\ \ \Longrightarrow \ \ \ x \leq 10;\]

\[Неравенство\ верно\ при:\]

\[3x - 14 \leq 0;\]

\[3x \leq 14;\]

\[x \leq 4\frac{2}{3};\]

\[Ответ:\ \ - 2,8 < x < 6.\]