Решебник по алгебре 8 класс Макарычев ФГОС Задание 410

 

\[\boxed{\text{410\ (410).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Внесем\ число\ под\ знак\ корня.\]

\[\textbf{а)}\ 7\sqrt{10} = \sqrt{7^{2} \cdot 10} =\]

\[= \sqrt{49 \cdot 10} = \sqrt{490}\]

\[\textbf{б)}\ 5\sqrt{3} = \sqrt{5^{2} \cdot 3} = \sqrt{25 \cdot 3} =\]

\[= \sqrt{75}\]

\[\textbf{в)}\ 6\sqrt{x} = \sqrt{6^{2} \cdot x} = \sqrt{36x}\]

\[\textbf{г)}\ 10\sqrt{y} = \sqrt{10^{2} \cdot y} = \sqrt{100y}\]

\[\textbf{д)}\ 3\sqrt{2a} = \sqrt{3^{2} \cdot 2a} = \sqrt{9 \cdot 2a} =\]

\[= \sqrt{18a}\]

\[\textbf{е)}\ 5\sqrt{3b} = \sqrt{5^{2} \cdot 3b} = \sqrt{25 \cdot 3b} =\]

\[= \sqrt{75b}\]

\[\boxed{\text{410.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\sqrt{a + \frac{a}{b}} = a\sqrt{\frac{a}{b}}\]

\[1.\ \sqrt{2\frac{2}{3}} = 2\sqrt{\frac{2}{3}}\text{\ \ \ \ \ \ \ \ \ \ \ }\sqrt{3\frac{3}{8}} = 3\sqrt{\frac{3}{8}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\text{\ \ \ }\sqrt{4\frac{4}{15}} = 4\sqrt{\frac{4}{15}}\text{\ \ \ \ \ \ \ }\sqrt{\frac{8}{3}} = \sqrt{\frac{4 \cdot 2}{3}}\text{\ \ \ \ \ \ \ }\]

\[\ \sqrt{\frac{24 + 3}{8}} = \sqrt{\frac{9 \cdot 3}{8}}\text{\ \ \ \ \ \ \ \ \ }\]

\[\text{\ \ }\sqrt{\frac{60 + 4}{15}} = \sqrt{\frac{16 \cdot 4}{15}}\text{\ \ \ \ \ }\]

\[\text{\ \ }\sqrt{\frac{8}{3}} = \sqrt{\frac{8}{3}}\text{\ \ \ \ \ \ \ \ \ }\sqrt{\frac{27}{8}} = \sqrt{\frac{27}{8}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \sqrt{\frac{64}{15}} = \sqrt{\frac{64}{15}}\]

\[2.\ \left( \sqrt{a + \frac{a}{b}} \right)^{2} = \left( a\sqrt{\frac{a}{b}} \right)^{2}\]

\[a + \frac{a}{b} = a^{2} \cdot \frac{a}{b}\]

\[\frac{ab + a}{b} = \frac{a^{3}}{b}\]

\[a^{2} = b + 1\]

\[b = a^{2} - 1\]

\[3.\ \]

\[b = 3^{2} - 1 = - 1 = 8\ \ \ \ \ \ \ \ \ \ \ \]

\[\ \sqrt{3\frac{3}{8}} = 3\sqrt{\frac{3}{8}}.\]

\[b = 5^{2} - 1 = 25 - 1 = 24\ \ \ \ \]

\[\ \sqrt{5\frac{5}{24}} = 5\sqrt{\frac{5}{24}}.\]