\[\boxed{\mathbf{553\ (553).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ \left( 2\sqrt{3} - 1 \right)\left( \sqrt{27} + 2 \right) =\]
\[= 2\sqrt{81} + 4\sqrt{3} - \sqrt{27} - 2 =\]
\[= 18 - 2 + 4\sqrt{3} - 3\sqrt{3} =\]
\[= 16 + \sqrt{3}\]
\[2)\ \left( \sqrt{5} - 2 \right)^{2} - \left( 3 + \sqrt{5} \right)^{2} =\]
\[= 5 - 4\sqrt{5} + 4 - 9 - 6\sqrt{5} - 5 =\]
\[= - 10\sqrt{5} - 5\ \]
\[3)\ \sqrt{\sqrt{17} - 4} \cdot \sqrt{\sqrt{17} + 4} =\]
\[= \sqrt{\left( \sqrt{17} - 4 \right)\left( \sqrt{17} + 4 \right)} =\]
\[= \sqrt{17 - 16} = 1\]
\[4)\ \left( 7 + 4\sqrt{3} \right)\left( 2 - \sqrt{3} \right)^{2} =\]
\[= \left( 7 + 4\sqrt{3} \right)\left( 7 - 4\sqrt{3} \right) =\]
\[= 49 - 16 \cdot 3 = 1\]
\[5)\ \left( \sqrt{6 + 2\sqrt{5}} - \sqrt{6 - 2\sqrt{5}} \right)^{2} =\]
\[= \left( \sqrt{\left( 6 + 2\sqrt{5} \right)\left( 6 - 2\sqrt{5} \right)} \right)^{2} =\]
\[= \sqrt{36 - 4 \cdot 5} = \sqrt{16} = 4\]
\[\boxed{\mathbf{5}\mathbf{5}\mathbf{3}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ 7\sqrt{2} = \sqrt{49 \cdot 2} = \sqrt{98};\]
\[2)\ 3\sqrt{13} = \sqrt{9 \cdot 13} = \sqrt{117};\]
\[3) - 2\sqrt{17} = - \sqrt{4} \cdot \sqrt{17} = - \sqrt{68};\]
\[4) - 10\sqrt{14} = - \sqrt{1400};\]
\[5)\ 5\sqrt{8} = \sqrt{25 \cdot 8} = \sqrt{200};\]
\[6)\ 6\sqrt{a} = \sqrt{36a};\]
\[7)\frac{1}{4}\sqrt{32} = \sqrt{\frac{1}{16} \cdot 32} = \sqrt{2};\]
\[8) - \frac{2}{3}\sqrt{54} = - \sqrt{\frac{4}{9} \cdot 54} = - \sqrt{24};\]
\[9)\frac{1}{8}\sqrt{128a} = \sqrt{\frac{1}{64} \cdot 128a} = \sqrt{2a};\]
\[10) - 0,3\sqrt{10b} = - \sqrt{\frac{9}{100} \cdot 10b} =\]
\[= - \sqrt{0,9b};\]
\[11)\ 3\ \sqrt{\frac{1}{3}} = \sqrt{9 \cdot \frac{1}{3}} = \sqrt{3};\]
\[\ 12)\frac{2}{9}\sqrt{\frac{27}{28}} = \sqrt{\frac{4}{81} \cdot \frac{27}{28}} = \sqrt{\frac{1}{21}}.\]