Решебник по алгебре 9 класс Мерзляк Задание 248

 

\[\boxed{\text{248\ (248).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ f(x) = \frac{x^{2} + 4x + 4}{x + 2}\]

\[x + 2 \neq 0\]

\[x \neq - 2.\]

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

\[y = \frac{x^{2} + 4x + 4}{x + 2}\]

\[y = \frac{(x + 2)^{2}}{x + 2}\]

\[y = x + 2\]

\[2)\ f(x) = \frac{x^{3}}{x};\ \ x \neq 0\]

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

\[y = \frac{x^{3}}{x}\]

\[y = x^{2}\]

\(\ \)

\[\boxed{\mathbf{248.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1) - 3x \geq 12\]

\[x \leq - 4.\]

\[2)\ \frac{1}{7}x < 4\]

\[x < 28.\]

\[3)\ \frac{5}{6}x > 30\]

\[x > 30 \cdot \frac{6}{5}\]

\[x > 25.\]

\[4) - 0,9x \geq 0\]

\[x \leq 0.\]

\[5)\ 9x - 2 > 25\]

\[9x > 27\]

\[x > 3.\]

\[6)\ 5 - x \leq 8\]

\[- x \leq 3\]

\[x \geq - 3.\]

\[7)\ 4 - 7x \geq 5\]

\[- 7x \geq 1\]

\[x \leq - \frac{1}{7}.\]

\[8)\ 42 - 4x < 2x\]

\[- 6x < - 42\]

\[x > 7.\]

\[9)\ \frac{x + 3}{8} > - 3\]

\[x + 3 > - 24\]

\[x > - 27.\]