ГДЗ по алгебре 8 класс Макарычев Задание 387

\[\boxed{\text{387\ (387).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \sqrt{10} \cdot \sqrt{40} = \sqrt{10 \cdot 40} =\]

\[= \sqrt{400} = 20\]

\[\textbf{б)}\ \sqrt{12} \cdot \sqrt{3} = \sqrt{12 \cdot 3} = \sqrt{36} = 6\]

\[\textbf{в)}\ \sqrt{162} \cdot \sqrt{2} = \sqrt{162 \cdot 2} =\]

\[= \sqrt{324} = 18\]

\[\textbf{г)}\ \sqrt{\frac{2}{3}} \cdot \sqrt{\frac{3}{8}} = \sqrt{\frac{2 \cdot 3}{3 \cdot 8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\]

\[\textbf{д)}\ \sqrt{110} \cdot \sqrt{4,4} = \sqrt{110 \cdot 4,4} =\]

\[= \sqrt{484} = 22\]

\[\textbf{е)}\ \sqrt{1\frac{4}{5}} \cdot \sqrt{0,2} = \sqrt{1,8 \cdot 0,2} =\]

\[= \sqrt{0,36} = 0,6\]

\[\textbf{ж)}\frac{\sqrt{999}}{\sqrt{111}} = \sqrt{\frac{999}{111}} = \sqrt{9} = 3\]

\[\textbf{з)}\frac{\sqrt{15}}{\sqrt{735}} = \sqrt{\frac{15}{735}} = \sqrt{\frac{1}{49}} = \frac{1}{7}\]