Решебник по алгебре 9 класс Мерзляк Задание 247

\[\boxed{\text{247\ (247).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ f(x) = \frac{x^{2} - 16}{x + 4}\]

\[x + 4 \neq 0\]

\[x \neq - 4\]

\[D(f) = ( - \infty; - 4) \cup ( - 4; + \infty).\]

\[\frac{x^{2} - 16}{x + 4} = \frac{(x + 4)(x - 4)}{(x + 4)} =\]

\[= x - 4\]

\[y = x - 4\]

\[2)\ f(x) = \frac{12x - 72}{x^{2} - 6x}\]

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

\[x \cdot (x - 6) \neq 0\]

\[x \neq 0\]

\[x \neq 6.\]

\[D(f) =\]

\[= ( - \infty;0) \cup (0;6) \cup (6; + \infty).\]

\[\frac{12x - 72}{x(x - 6)} = \frac{12 \cdot (x - 6)}{x \cdot (x - 6)} = \frac{12}{x}\]

\[y = \frac{12}{x}\]

\[3)\ f(x) = \frac{x^{2} - 9}{x^{2} - 9}\]

\[x^{2} - 9 \neq 0\]

\[x \neq 3\]

\[x \neq - 3.\]

\[D(f) =\]

\[= ( - \infty; - 3) \cup ( - 3;3) \cup (3; + \infty).\]

\[\frac{x^{2} - 9}{x^{2} - 9} = 1\]

\[y = 1\]

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