Решебник по алгебре 9 класс Макарычев Задание 262

\[\boxed{\text{262}\text{\ (262)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y = \sqrt{0,1x - 2}\]

\[0,1x - 2 \geq 0\]

\[0,1x \geq 2\]

\[x \geq 20.\]

\[\textbf{б)}\ y = \sqrt[4]{5 - 2x}\]

\[5 - 2x \geq 0\]

\[2x \leq 5\]

\[x \leq 2,5.\]

\[\textbf{в)}\ y = \sqrt[3]{8x + 1}\]

\[x \in ( - \infty;\ + \infty)\text{.\ }\]

\[\boxed{\text{262.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{2ab + 2by + ay + y^{2}}{2ab - 2by + ay - y^{2}} =\]

\[= \frac{a \cdot (2b + y) + y(2b + y)}{a(2b + y) - y(2b + y)} =\]

\[= \frac{(2b + y)(a + y)}{(2b + y)(a - y)} =\]

\[= \frac{a + y}{a - y}.\]

\[(В\ учебнике\ опечатка).\]

\[\textbf{б)}\ \frac{9x^{2} + 6x + 4}{27x^{3} - 8} =\]

\[= \frac{9x^{2} + 6x + 4}{(3x - 2)\left( 9x^{2} + 6x + 4 \right)} =\]

\[= \frac{1}{3x - 2}.\]