ГДЗ по алгебре 11 класс Никольский Параграф 1. Функции и их графики Задание 8

\[\boxed{\mathbf{8.}}\]

\[\textbf{а)}\ y = \sqrt{x - 1}\]

\[x - 1 \geq 0\]

\[x \geq 1.\]

\[D(f) = \lbrack 1; + \infty).\]

\[\textbf{б)}\ y = \sqrt[3]{x + 1}\]

\[D(f) = ( - \infty; + \infty).\]

\[\textbf{в)}\ y = \sqrt{x^{2} - 1}\]

\[x^{2} - 1 \geq 0\]

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

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

\[D(f) = ( - \infty; - 1\rbrack \cup \lbrack 1; + \infty).\]

\[\textbf{г)}\ y = \frac{x^{2} - 9}{x^{2} - 4}\]

\[x^{2} - 4 \neq 0\]

\[x^{2} \neq 4\]

\[x \neq \pm 2.\]

\[D(f) =\]

\[= ( - \infty; - 2) \cup ( - 2;2) \cup (2; + \infty).\]

\[\textbf{д)}\ y = \frac{1}{\sqrt{x^{2} - x}}\]

\[x^{2} - x > 0\]

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

\[x < 0;\ \ x > 1.\]

\[D(f) = ( - \infty;0) \cup (1; + \infty).\]

\[\textbf{е)}\ y = \frac{\sqrt{x^{2} + x}}{x + 4}\]

\[x + 4 \neq 0\]

\[x \neq - 4.\]

\[x^{2} + x \geq 0\]

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

\[x \leq - 1;\ \ x \geq 0.\]

\[D(f) =\]

\[= ( - \infty; - 4) \cup ( - 4; - 1\rbrack \cup \lbrack 0; + \infty).\]