ГДЗ по алгебре 8 класс Макарычев Задание 417

\[\boxed{\text{417\ (417).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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

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

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

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

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

\[a(b + a) = a^{3}\]

\[b + 1 = a^{2} - искомое\ \]

\[соотношение.\]

\[1)\ \sqrt{2\frac{2}{3}} = 2\sqrt{\frac{2}{3}}\ \]

\[\left( \sqrt{2\frac{2}{3}} \right)^{2} = \left( 2\sqrt{\frac{2}{3}} \right)^{2}\]

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

\[2\frac{2}{3} = 4 \cdot \frac{2}{3}\]

\[2\frac{2}{3} = \frac{8}{3}\]

\[2\frac{2}{3} = 2\frac{2}{3}.\]

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

\[\left( \sqrt{3\frac{3}{8}} \right)^{2} = \left( 3\sqrt{\frac{3}{8}}\ \right)^{2}\]

\[3\frac{3}{8} = 9 \cdot \frac{3}{8}\]

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

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

\[\sqrt{4\frac{4}{15}} = 4\sqrt{\frac{4}{15}}\]

\[\left( \sqrt{4\frac{4}{15}} \right)^{2} = \left( 4\sqrt{\frac{4}{15}} \right)^{2}\]

\[4\frac{4}{15} = 16 \cdot \frac{4}{15}\]

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

\[3)\ \mathbf{a = 3}:\ \ \ \ \ \ \ \ \ \ \]

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

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

\[\mathbf{a = 5}:\ \ \ \ \ \ \ \ \ \ \ \ \]

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

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