\[\boxed{\text{491\ (491).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ y = x^{2} - 9\ \]
\[y > x^{2} - 9;\]
\[\textbf{б)}\ y = (x + 2)^{2}\]
\[y < (x + 2)^{2}.\]
\[\boxed{\text{491.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\left\{ \begin{matrix} x^{2} + y^{2} = 5 \\ x - y = m\ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + m\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 2my + m^{2} + y^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + m\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2y² + 2m + m² - 5 = 0 \\ \end{matrix} \right.\ \]
\[\textbf{а)}\ D = m^{2} - 2 \cdot \left( m^{2} - 5 \right) =\]
\[= - m^{2} + 10.\]
\[Система\ имеет\ 1\ решение\ при\ \]
\[D = 0 \Longrightarrow - m^{2} + 10 = 0 \Longrightarrow\]
\[\Longrightarrow m^{2} = 10 \Longrightarrow m = \pm \sqrt{10}.\]
\[\textbf{б)}\ Система\ имеет\ 2\ решения\ \]
\[при\ D > 0 \Longrightarrow - m^{2} + 10 > 0 \Longrightarrow\]
\[\Longrightarrow m^{2} - 10 < 0 \Longrightarrow\]
\[\Longrightarrow \left( m - \sqrt{10} \right)\left( m + \sqrt{10} \right) < 0 \Longrightarrow\]
\[\Longrightarrow m \in \left( - \sqrt{10};\sqrt{10} \right).\]
\[Ответ:а)\ при\ m = - \sqrt{10},\]
\[при\ m = \sqrt{10};\]
\(б)\ при\ \ m \in \left( - \sqrt{10};\sqrt{10} \right).\)