\[\boxed{\text{2\ (2).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\frac{c^{2} - 4c}{2c + 1}\text{\ \ }\]
\[если\ c = - 3:\]
\[\frac{c^{2} - 4c}{2c + 1} = \frac{( - 3)^{2} - 4 \cdot ( - 3)}{2 \cdot ( - 3) + 1} =\]
\[= \frac{9 + 12}{- 6 + 1} = - \frac{21}{5} = - 4,2.\]
\[2)\ \frac{c^{2} - 4c}{2c + 1}\ \]
\[если\ c = 0:\]
\[\frac{c^{2} - 4c}{2c + 1} = \frac{0 - 4 \cdot 0}{2 \cdot 0 + 1} = 0.\]