\[\boxed{\text{385\ (385).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \sqrt{2} \cdot \sqrt{8} = \sqrt{2 \cdot 8} = \sqrt{16} = 4\]
\[\textbf{б)}\ \sqrt{27} \cdot \sqrt{3} = \sqrt{27 \cdot 3} = \sqrt{81} =\]
\[= 9\]
\[\textbf{в)}\ \sqrt{28} \cdot \sqrt{7} = \sqrt{28 \cdot 7} = \sqrt{196} =\]
\[= 14\]
\[\textbf{г)}\ \sqrt{2} \cdot \sqrt{32} = \sqrt{2 \cdot 32} = \sqrt{64} = 8\]
\[\textbf{д)}\ \sqrt{13} \cdot \sqrt{52} = \sqrt{13 \cdot 52} =\]
\[= \sqrt{13 \cdot 13 \cdot 4} = \sqrt{13^{2} \cdot 2^{2}} =\]
\[= 13 \cdot 2 = 26\]
\[\textbf{е)}\ \sqrt{63 \cdot 7} = \sqrt{63 \cdot 7} = \sqrt{441} =\]
\[= 21\]
\[\textbf{ж)}\ \sqrt{50} \cdot \sqrt{4,5} = \sqrt{50 \cdot 4,5} =\]
\[= \sqrt{225} = 15\]
\[\textbf{з)}\ \sqrt{1,2} \cdot \sqrt{3\frac{1}{3}} = \sqrt{\frac{12}{10} \cdot \frac{10}{3}} =\]
\[= \sqrt{\frac{12}{3}} = \sqrt{4} = 2\]