\[\boxed{\text{248\ (248).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ f(x) = \frac{x^{2} + 4x + 4}{x + 2}\]
\[x + 2 \neq 0\]
\[x \neq - 2.\]
\[D(f) = ( - \infty; - 2) \cup ( - 2; + \infty).\]
\[y = \frac{x^{2} + 4x + 4}{x + 2}\]
\[y = \frac{(x + 2)^{2}}{x + 2}\]
\[y = x + 2\]
\[x\] | \[0\] | \[1\] |
---|---|---|
\[y\] | \[2\] | \[3\] |
\[2)\ f(x) = \frac{x^{3}}{x};\ \ x \neq 0\]
\[D(f) = ( - \infty;0) \cup (0; + \infty).\]
\[y = \frac{x^{3}}{x}\]
\[y = x^{2}\]
\[x\] | \[1\] | \[2\] | \[- 1\] | \[- 2\] |
---|---|---|---|---|
\[y\] | \[1\] | \[4\] | \[1\] | \[4\] |
\(\ \)