ГДЗ по алгебре 8 класс Макарычев Задание 412

\[\boxed{\text{412\ (412).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 3\sqrt{\frac{1}{3}} = \sqrt{3^{2}} \cdot \sqrt{\frac{1}{3}} = \ \sqrt{9 \cdot \frac{1}{3}} =\]

\[= \sqrt{3}\]

\[\textbf{б)}\ 2\sqrt{\frac{3}{4}} = \sqrt{2^{2}} \cdot \sqrt{\frac{3}{4}} = \sqrt{4 \cdot \frac{3}{4}} =\]

\[= \sqrt{3}\]

\[\textbf{в)}\ \frac{1}{3}\sqrt{18} = \sqrt{\left( \frac{1}{3} \right)^{2}} \cdot \sqrt{18} =\]

\[= \sqrt{\frac{1}{9} \cdot 18} = \sqrt{2}\ \]

\[\textbf{г)} - 10\sqrt{0,02} =\]

\[= - \sqrt{10^{2}} \cdot \sqrt{0,02} =\]

\[= - \sqrt{100 \cdot 0,02} = - \sqrt{2}\]

\[\textbf{д)}\ 5\sqrt{\frac{a}{5}\ } = \sqrt{5^{2}} \cdot \sqrt{\frac{a}{5}} = \sqrt{25 \cdot \frac{a}{5}} =\]

\[= \sqrt{5a}\]

\[\textbf{е)} - \frac{1}{2}\sqrt{12x} = - \sqrt{\left( \frac{1}{2} \right)^{2}} \cdot \sqrt{12x} =\]

\[= - \sqrt{\frac{1}{4} \cdot 12x} = - \sqrt{3x}\]

\[\textbf{ж)} - 0,1\sqrt{1,2a} =\]

\[= - \sqrt{{0,1}^{2}} \cdot \sqrt{1,2a} =\]

\[= - \sqrt{0,01 \cdot 1,2a} = - \sqrt{0,012a}\]

\[\textbf{з)} - \frac{1}{3}\sqrt{0,9a} = - \sqrt{\left( \frac{1}{3} \right)^{2}} \cdot \sqrt{0,9a} =\]

\[= - \sqrt{\frac{1}{9} \cdot \frac{9}{10}a} = - \sqrt{\frac{1}{10}a} =\]

\[= - \sqrt{0,1a}\]

\[\textbf{и)} - 6\sqrt{6b} = - \sqrt{6^{2}} \cdot \sqrt{6b} =\]

\[= - \sqrt{36 \cdot 6b} = - \sqrt{216b}\ \]