Решебник по алгебре 8 класс Макарычев ФГОС Задание 417

 

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

Пояснение.

Решение.

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

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

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

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

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

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

\[b + 1 = a^{2} - искомое\ \]

\[соотношение.\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\boxed{\text{417.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

При любом a, при котором выражение √a имеет смысл, верно равенство:

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

Формулы сокращенного умножения:

\[(m + n)^{2} = m^{2} + 2mn + n^{2};\]

\[(m - n)^{2} = m^{2} - 2mn + n^{2};\]

\[(m - n)(m + n) = m^{2} - n^{2}.\]

Решение.

\[\textbf{а)}\ \left( 2\sqrt{5} + 1 \right)\left( 2\sqrt{5} - 1 \right) =\]

\[= \left( 2\sqrt{5} \right)^{2} - 1 = 4 \cdot 5 - 1 =\]

\[= 20 - 1 = 19\]

\[\textbf{б)}\ \left( 5\sqrt{7} - \sqrt{13} \right)\left( 5\sqrt{7} + \sqrt{13} \right) =\]

\[= \left( 5\sqrt{7} \right)^{2} - \left( \sqrt{13} \right)^{2} =\]

\[= 25 \cdot 7 - 13 = 175 - 13 = 162\]

\[\textbf{в)}\ \left( 3\sqrt{2} - 2\sqrt{3} \right)\left( 2\sqrt{3} + 3\sqrt{2} \right) =\]

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

\[= 9 \cdot 2 - 4 \cdot 3 = 18 - 12 = 6\]

\[\textbf{г)}\ \left( 1 + 3\sqrt{5} \right)^{2} =\]

\[= 1 + 2 \cdot 1 \cdot 3\sqrt{5} + \left( 3\sqrt{5} \right)^{2} =\]

\[= 1 + 6\sqrt{5} + 9 \cdot 5 = 46 + 6\sqrt{5}\]

\[\textbf{д)}\ \left( 2\sqrt{3} - 7 \right)^{2} =\]

\[= \left( 2\sqrt{3} \right)^{2} - 2 \cdot 2\sqrt{3} \cdot 7 + 7^{2} =\]

\[= 4 \cdot 3 - 28\sqrt{3} + 49 =\]

\[= 61 - 28\sqrt{3}\]

\[\textbf{е)}\ \left( 2\sqrt{10} - \sqrt{2} \right)^{2} =\]

\[= \left( 2\sqrt{10} \right)^{2} - 2 \cdot 2\sqrt{10} \cdot \sqrt{2} + \left( \sqrt{2} \right)^{2} =\]

\[= 4 \cdot 10 - 4\sqrt{20} + 2 = \ 42 - 8\sqrt{5}\]