\[y = \sqrt{7x - 2x^{2}} + \frac{2x + 3}{9 - x^{2}}\]
\[9 - x^{2} \neq 0\]
\[x^{2} \neq 9\]
\[x \neq \pm 3.\]
\[7x - 2x^{2} \geq 0\]
\[2x^{2} - 7x \leq 0\]
\[2x(x - 3,5) \leq 0\]
\[0 \leq x \leq 3,5.\]
\[Ответ:x \in \lbrack 0;3) \cup (3;3,5\rbrack.\]