\[\boxed{\text{378}\text{\ (378)}\text{.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[y = x^{2} + px + q;\ \ \ \]
\[\text{A\ }(2;5) - вершина:\]
\[\left\{ \begin{matrix} - \frac{p}{2} = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + px + q = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\text{\ \ }\left\{ \begin{matrix} p = - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 4x + q = 5 \\ \end{matrix} \right.\ \]
\[x^{2} - 4x + q - 5 = 0\]
\[D = 16 - 4q + 20 = 36 - 4q\]
\[36 - 4q = 0\]
\[q = 9\]
\[Ответ:p = - 4,\ \ \ q = 9.\]
\[\boxed{\text{378.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ \left( \sqrt{11} + \sqrt{6} \right)\left( \sqrt{11} - \sqrt{6} \right) =\]
\[= 11 - 6 = 5\]
\[2)\ \left( \sqrt{32} - 5 \right)\left( \sqrt{32} + 5 \right) =\]
\[= 32 - 25 = 7\]
\[3)\ \left( \sqrt{5} + \sqrt{3} \right)^{2} = 5 + 2\sqrt{15} + 3 =\]
\[= 8 + 2\sqrt{15}\]
\[4)\ \left( \sqrt{10} + 8 \right)^{2} =\]
\[= 10 + 16\sqrt{10} + 64 =\]
\[= 74 + 16\sqrt{10}\ \]