\[\boxed{\text{445\ (445).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \sqrt{11 + 6\sqrt{2}} - \sqrt{2} =\]
\[= \sqrt{9 + 2 + 6\sqrt{2}} - \sqrt{2} =\]
\[= \sqrt{3^{2} + 6\sqrt{2} + \left( \sqrt{2} \right)^{2}} =\]
\[= \sqrt{\left( 3 + \sqrt{2} \right)^{2}} - \sqrt{2} =\]
\[= \left| 3 + \sqrt{2} \right| - \sqrt{2} =\]
\[= 3 + \sqrt{2} - \sqrt{2} =\]
\[= 3\ \left( так\ как\ 3 + \sqrt{2} > 0 \right).\]
\[\textbf{б)}\ \sqrt{27 - 5\sqrt{8}} + \sqrt{2} =\]
\[= \sqrt{27 - 5 \cdot 2\sqrt{2}} + \sqrt{2} =\]
\[= \sqrt{27 - 10\sqrt{2}} + \sqrt{2} =\]
\[= \sqrt{25 + 2 - 10\sqrt{2}} + \sqrt{2} =\]
\[= \sqrt{5^{2} - 2 \cdot 5\sqrt{2} + \left( \sqrt{2} \right)^{2}} =\]
\[= \sqrt{\left( 5 - \sqrt{2} \right)^{2}} + \sqrt{2} =\]
\[= \left| 5 - \sqrt{2} \right| + \sqrt{2} =\]
\[= 5 - \sqrt{2} + \sqrt{2} =\]
\[= 5\ \left( так\ как\ 5 = \sqrt{25} > \sqrt{2} \right).\]
\[\boxed{\text{445.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
Пояснение.
Подкоренные выражения приведем к общему знаменателю.
Формулы сокращенного умножения:
\[(a + b)^{2} = a^{2} + 2ab + b^{2};\]
\[(a - b)^{2} = a^{2} - 2ab + b^{2}.\]
Решение.
\[\textbf{а)}\ b \geq 1:\]
\[\ \sqrt{\frac{b + 1}{2} - {\sqrt{b}}^{\backslash 2}} - \sqrt{\frac{b + 1}{2} + {\sqrt{b}}^{\backslash 2}} =\]
\[= \sqrt{\frac{b + 1 - 2\sqrt{b}}{2}} - \sqrt{\frac{b + 1 + 2\sqrt{b}}{2}} =\]
\[= \sqrt{\frac{\left( \sqrt{b} - 1 \right)^{2}}{2}} - \sqrt{\frac{\left( \sqrt{b} + 1 \right)^{2}}{2}} =\]
\[= \frac{\sqrt{b} - 1}{\sqrt{2}} - \frac{\sqrt{b} + 1}{\sqrt{2}} =\]
\[= \frac{\sqrt{b} - 1 - \sqrt{b} - 1}{\sqrt{2}} =\]
\[= - \frac{2}{\sqrt{2}} = \frac{- 2\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = - \frac{2\sqrt{2}}{2} =\]
\[= - \sqrt{2}\]
\[\textbf{б)}\ c \geq 4\ \]
\[\sqrt{\frac{c + 4}{4} + {\sqrt{c}}^{\backslash 4}} - \sqrt{\frac{c + 4}{4} - {\sqrt{c}}^{\backslash 4}} =\]
\[= \sqrt{\frac{c + 4 + 4\sqrt{c}}{4}} - \sqrt{\frac{c + 4 - 4\sqrt{c}}{4}} =\]
\[= \sqrt{\frac{\left( \sqrt{c} + 2 \right)^{2}}{4}} - \sqrt{\frac{\left( \sqrt{c} - 2 \right)^{2}}{4}} =\]
\[= \frac{\sqrt{c} + 2}{2} - \frac{\sqrt{c} - 2}{2} =\]
\[= \frac{\sqrt{c} + 2 - \sqrt{c} + 2}{2} = \frac{4}{2} = 2\]