Решебник по алгебре 8 класс Мерзляк ФГОС Задание 526

 

\[\boxed{\mathbf{526\ (526).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3};\]

\[2)\ \sqrt{24} = 2\sqrt{6};\]

\[3)\ \sqrt{20} = 2\sqrt{5};\]

\[4)\ \sqrt{125} = 5\sqrt{5};\]

\[5)\frac{1}{8}\sqrt{96} = \frac{1}{8} \cdot 4\sqrt{6} = \frac{\sqrt{6}}{2};\]

\[6)\ 0,4\sqrt{250} = 0,4 \cdot 5\sqrt{10} =\]

\[= 2\sqrt{10};\]

\[7) - 2\sqrt{0,18} = - 2 \cdot 3\sqrt{0,02} =\]

\[= - 6\sqrt{0,02} = - 0,6\sqrt{2};\]

\[8)\frac{4}{9}\sqrt{63} = \frac{4}{9} \cdot 3\sqrt{7} = \frac{4\sqrt{7}}{3};\]

\[9)\ 0,8\sqrt{1250} = 0,8 \cdot 25\sqrt{2} =\]

\[= 20\sqrt{2};\]

\[10)\frac{3}{7}\sqrt{98} = \frac{3}{7} \cdot 7\sqrt{2} = 3\sqrt{2};\]

\[11)\ 10\sqrt{0,03} = \sqrt{3};\]

\[12)\ 0,7\sqrt{1000} = 7\sqrt{10}\text{.\ }\]

\[\boxed{\text{526.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ \sqrt{12} \cdot \sqrt{3} = \sqrt{36} = 6.\]

\[2)\ \sqrt{32} \cdot \sqrt{2} = \sqrt{64} = 8.\]

\[3)\ \sqrt{18} \cdot \sqrt{50} = \sqrt{900} = 30.\]

\[4)\ \sqrt{0,009} \cdot \sqrt{1000} = \sqrt{9} = 3.\]

\[5)\ \sqrt{200} \cdot \sqrt{0,18} = \sqrt{36} = 6.\]

\[6)\ \sqrt{13} \cdot \sqrt{2} \cdot \sqrt{26} = \sqrt{676} = 26.\]

\[7)\ \sqrt{2,4} \cdot \sqrt{1\frac{2}{3}} = \sqrt{2\frac{4}{10}} \cdot \sqrt{1\frac{2}{3}} =\]

\[= \sqrt{\frac{24 \cdot 5}{10 \cdot 3}} = \sqrt{\frac{8 \cdot 1}{2 \cdot 1}} = \sqrt{4} = 2.\]

\[8)\ \sqrt{\frac{2}{11}} \cdot \sqrt{8} \cdot \sqrt{\frac{1}{11}} = \sqrt{\frac{2 \cdot 8}{11 \cdot 11}} =\]

\[= \sqrt{\frac{16}{121}} = \frac{4}{11}.\]

\[9)\ \sqrt{2^{3} \cdot 3} \cdot \sqrt{2^{5} \cdot 3^{3}} = \sqrt{2^{8} \cdot 3^{4}} =\]

\[= 2^{4} \cdot 3^{2} = 16 \cdot 9 = 144.\ \]