Решебник по алгебре 9 класс Макарычев Задание 263

\[\boxed{\text{263}\text{\ (263)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = x;\ \ y = \sqrt{x};\ \ y = \sqrt[3]{x}\]

\[\textbf{а)}\ \sqrt{x} = x \Longrightarrow x_{1} = 0\ \ и\ \ x_{2} = 1;\]

\[\sqrt{x} < x \Longrightarrow x > 1;\]

\[\sqrt{x} > x \Longrightarrow 0 < x < 1.\]

\[\textbf{б)}\ \sqrt[3]{x} = x \Longrightarrow x_{1} = 1\ \ и\ \ x_{2} =\]

\[= 0,\ x_{3} = - 1;\]

\[\sqrt[3]{x} < x \Longrightarrow x^{3} < x^{3} \Longrightarrow\]

\[\Longrightarrow x\left( x^{2} - 1 \right) > 0 \Longrightarrow\]

\[\Longrightarrow x \in ( - 1;0) \cup (1;\ + \infty);\]

\[\sqrt[3]{x} > x \Longrightarrow x > x^{3} \Longrightarrow\]

\[\Longrightarrow x\left( x^{2} - 1 \right) < 0 \Longrightarrow\]

\[\Longrightarrow x \in ( - \infty;\ - 1) \cup (0;1).\]

\[\boxed{\text{263.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

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

\[x^{2} = x + 6\]

\[y = x^{2};\ \ \ y = x + 6.\]

\[x = - 2;\ \ x = 3.\]

\[Ответ:x = - 2;\ \ x = 3.\]