ГДЗ по алгебре 8 класс Макарычев Задание 401

\[\boxed{\text{401\ (401).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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\[степеней:\]

\[\mathbf{a}^{\mathbf{\text{mn}}}\mathbf{=}\left( \mathbf{a}^{\mathbf{m}} \right)^{\mathbf{n}}\mathbf{.}\]

\[\textbf{а)}\ \sqrt{2^{4}} = \sqrt{\left( 2^{2} \right)^{2}} = \left| 2^{2} \right| = 2^{2} =\]

\[= 4\]

\[\textbf{б)}\ \sqrt{3^{4}} = \sqrt{\left( 3^{2} \right)^{2}} = \left| 3^{2} \right| = 3^{2} =\]

\[= 9\]

\[\textbf{в)}\ \sqrt{2^{6}} = \sqrt{\left( 2^{3} \right)^{2}} = \left| 2^{3} \right| = |8| =\]

\[= 8\]

\[\textbf{г)}\ \sqrt{10^{8}} = \sqrt{\left( 10^{4} \right)^{2}} = \left| 10^{4} \right| =\]

\[= 10^{4} = 10\ 000\]

\[\textbf{д)}\ \sqrt{( - 5)^{4}} = \sqrt{\left( ( - 5)^{2} \right)^{2}} =\]

\[= {|( - 5)}^{2}| = 25\]

\[\textbf{е)}\ \sqrt{( - 2)^{8}} = \sqrt{\left( ( - 2)^{4} \right)^{2}} =\]

\[= |( - 2)^{4}| = 16\]

\[\textbf{ж)}\sqrt{3^{4}\ \cdot 5^{2}} = \sqrt{\left( 3^{2} \cdot 5 \right)^{2}} =\]

\[= \left| 3^{2} \cdot 5 \right| = 9 \cdot 5 = 45\]

\[\textbf{з)}\ \sqrt{2^{6} \cdot 7^{4}} = \sqrt{\left( 2^{3} \cdot 7^{2} \right)^{2}} =\]

\[= |2^{3} \cdot 7^{2}| = 8 \cdot 49 = 392\ \]