ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1031

\[1)\ y = \sqrt{\frac{x - 3}{x + 3}}\]

\[\frac{x - 3}{x + 3} \geq 0\]

\[x < - 3;\ \ \ x \geq 3.\]

\[x \in ( - \infty; - 3) \cup \lbrack 3; + \infty).\]

\[2)\ y = \sqrt{4 - \frac{9}{x + 1} + \frac{1}{x - 3}}\]

\[4 - \frac{9}{x + 1} + \frac{1}{x - 3} \geq 0\]

\[\frac{4(x + 1)(x - 3) - 9(x - 3) + (x + 1)}{(x + 1)(x - 3)} \geq 0\]

\[\frac{4x^{2} - 12x + 4x - 12 - 9x + 27 + x + 1}{(x + 1)(x - 3)} \geq 0\]

\[\frac{4x^{2} - 16x + 16}{(x + 1)(x - 3)} \geq 0\]

\[\frac{4\left( x^{2} - 4x + 4 \right)}{(x + 1)(x - 3)} \geq 0\]

\[\frac{4(x - 2)^{2}}{(x + 1)(x - 3)} \geq 0\]

\[(x + 1)(x - 3) > 0\]

\[x < - 1;\ \ \ x > 3.\]

\[x - 2 = 0\]

\[x = 2.\]

\[x \in ( - \infty;\ - 1) \cup \left\{ 2 \right\} \cup (3;\ + \infty).\]

\[3)\ y = \sqrt{\frac{x^{2} - 6x - 16}{x^{2} - 12x + 11}}\]

\[x^{2} - 6x - 16 = 0\]

\[D = 36 + 64 = 100\]

\[x_{1} = \frac{6 - 10}{2} = - 2;\]

\[x_{2} = \frac{6 + 10}{2} = 8.\]

\[x^{2} - 12x + 11 = 0\]

\[D = 144 - 44 = 100\]

\[x_{1} = \frac{12 - 10}{2} = 1;\]

\[x_{2} = \frac{12 + 10}{2} = 11.\]

\[Область\ определения:\]

\[\frac{x^{2} - 6x - 16}{x^{2} - 12x + 11} \geq 0\]

\[\frac{(x + 2)(x - 8)}{(x - 1)(x - 11)} \geq 0\ \]

\[x \leq - 2;\]

\[1 < x \leq 8;\]

\[x > 11.\]

\[x \in ( - \infty;\ - 2\rbrack \cup (1;\ 8\rbrack \cup (11;\ + \infty).\]