ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 634

\[\boxed{\mathbf{634}.}\]

\[1)\ \sqrt{x - 1} < a\]

\[\sqrt{x - 1} \geq 0 \Longrightarrow a \geq 0;\]

\[при\ a < 0 - решений\ нет;\]

\[при\ a = 0:\]

\[\sqrt{x - 1} = 0\]

\[x - 1 = 0\]

\[x = 1.\]

\[при\ a > 0:\]

\[\left( \sqrt{x - 1} \right)^{2} < a^{2}\]

\[x - 1 < a^{2}\]

\[x < a^{2} + 1;\ \ \ x \geq 1.\]

\[Ответ:\ \]

\[при\ a < 0 - решений\ нет;\]

\[при\ a = 0:x = 1;\]

\[при\ a > 0:\ \ 1 \leq x \leq a^{2} + 1.\]

\[2)\ \sqrt{2ax - x^{2}} \geq a - x;\ \ \ a \leq 0.\]

\[2ax - x^{2} = (a - x)^{2}\]

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

\[2x^{2} - 4ax + a^{2} = 0\]

\[D_{1} = 4a^{2} - 2a^{2} = 2a^{2}\]

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

\[\ x_{2} = \frac{2a - a\sqrt{2}}{2} = a - \frac{\sqrt{2}}{2}\text{a.}\]

\[2\left( x - x_{1} \right)\left( x - x_{2} \right) \leq 0;\ \ \ \]

\[x < a;\ \ a \leq 0.\]

\[2ax - x^{2} \geq 0;\ \ x > a\]

\[- x(x - 2a) \geq 0\]

\[x(x - 2a) \leq 0.\]

\[Ответ:\ \ a + \frac{\sqrt{2}}{2}a \leq x \leq 0.\]