\[\boxed{\text{440\ (440).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\mathbf{Воспользуемся\ формулами:}\]
\[\mathbf{a}^{\mathbf{2}}\mathbf{-}\mathbf{b}^{\mathbf{2}}\mathbf{=}\left( \mathbf{a - b} \right)\left( \mathbf{a + b} \right)\mathbf{;}\]
\[\mathbf{a}^{\mathbf{2}}\mathbf{+ 2}\mathbf{ab +}\mathbf{b}^{\mathbf{2}}\mathbf{=}\left( \mathbf{a + b} \right)^{\mathbf{2}}\mathbf{.}\]
Решение.
\[\frac{9 - x^{2}}{\mathbf{4}\mathbf{x}} \cdot \frac{\mathbf{8}\mathbf{x}}{x^{2} + 6x + 9} - 2 =\]
\[= \frac{(3 - x)(3 + x) \cdot 2}{(x + 3)^{2}} - 2 =\]
\[= \frac{2(3 - x) - 2(x + 3)}{x + 3} =\]
\[= \frac{6 - 2x - 2x - 6}{x + 3} = \frac{- 4x}{x + 3}\]
\[При\ x = - 2,5:\]
\[- \frac{4x}{x + 3} = - \frac{4 \cdot ( - 2,5)}{- 2,5 + 3} = \frac{10}{0,5} =\]
\[= \frac{100}{5} = 20.\]