\[\boxed{\mathbf{709\ (709).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ y = \frac{\sqrt{2 - x}}{x + 2}\]
\[\left\{ \begin{matrix} 2 - x \geq 0 \\ x + 2 \neq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x \leq 2\ \ \ \\ x \neq - 2 \\ \end{matrix} \right.\ \]
\[Ответ:x \in ( - \infty;\ - 2) \cup ( - 2;2\rbrack.\]
\[2)\ y = \frac{\sqrt{6 - 5x - x^{2}}}{x - 1}\]
\[\left\{ \begin{matrix} 6 - 5x - x^{2} \geq 0 \\ x - 1 \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\text{\ \ }\left\{ \begin{matrix} x² + 5x - 6 \leq 0 \\ x \neq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x^{2} + 5x - 6 = 0\]
\[x_{1} + x_{2} = - 5,\ \ x_{1} = - 6\]
\[x_{1}x_{2} = - 6,\ \ x_{2} = 1\]
\[Ответ:x \in \lbrack - 6;1).\]