Решебник по алгебре 8 класс Мерзляк ФГОС Задание 361

 

\[\boxed{\text{36}\text{1}\text{\ (361).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ y = \frac{x^{3} + x^{2}}{x + 1}\]

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

\[y = x^{2};\ \ при\ x \neq - 1\]

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

\[y = \frac{x^{2}\left( x^{2} - 4 \right)}{x^{2} - 4} = x^{2}\]

\[y = x^{2};\ \ при\ x \neq 2,;\ x \neq - 2\ \ \]

\[\boxed{\text{361.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[f(x) =\]

\[= \left\{ \begin{matrix} 4,\ \ если\ x \leq - 2\ \ \ \ \ \ \ \ \ \ \ \\ x^{2},\ \ если - 2 < x < 1 \\ 2x - 1,\ \ если\ x \geq 1\ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ f( - 3) = 4\]

\[f( - 2) = 4\]

\[f( - 1) = x^{2} = ( - 1)^{2} = 1\]

\[f(1) = 2x - 1 = 2 \cdot 1 - 1 = 1\]

\[f(3) = 2x - 1 = 2 \cdot 3 - 1 = 5\]

\[f(0,5) = x^{2} = {0,5}^{2} = 0,25\]

\[2)\ f(x) = x^{2}\]

\[f(x) = 2x - 1\]

\[3)\ При\ 0 < a < 4.\]