Решебник по алгебре 8 класс Мерзляк Задание 527

\[\boxed{\mathbf{527\ (527).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 7\sqrt{2} = \sqrt{49 \cdot 2} = \sqrt{98};\]

\[2)\ 3\sqrt{13} = \sqrt{9 \cdot 13} = \sqrt{117};\]

\[3) - 2\sqrt{17} = - \sqrt{4} \cdot \sqrt{17} = - \sqrt{68};\]

\[4) - 10\sqrt{14} = - \sqrt{1400};\]

\[5)\ 5\sqrt{8} = \sqrt{25 \cdot 8} = \sqrt{200};\]

\[6)\ 6\sqrt{a} = \sqrt{36a};\]

\[7)\frac{1}{4}\sqrt{32} = \sqrt{\frac{1}{16} \cdot 32} = \sqrt{2};\]

\[8) - \frac{2}{3}\sqrt{54} = - \sqrt{\frac{4}{9} \cdot 54} = - \sqrt{24};\]

\[9)\frac{1}{8}\sqrt{128a} = \sqrt{\frac{1}{64} \cdot 128a} = \sqrt{2a};\]

\[10) - 0,3\sqrt{10b} = - \sqrt{\frac{9}{100} \cdot 10b} =\]

\[= - \sqrt{0,9b};\]

\[11)\ 3\ \sqrt{\frac{1}{3}} = \sqrt{9 \cdot \frac{1}{3}} = \sqrt{3};\]

\[\ 12)\frac{2}{9}\sqrt{\frac{27}{28}} = \sqrt{\frac{4}{81} \cdot \frac{27}{28}} = \sqrt{\frac{1}{21}}.\]