Решебник по алгебре 8 класс Макарычев ФГОС Задание 424

 

\[\boxed{\text{424\ (424).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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

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

\[= 20 - 1 = 19\]

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

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

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

\[= 175 - 13 = 162\]

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

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

\[= 9 \cdot 2 - 4 \cdot 3 =\]

\[= 18 - 12 = 6\]

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

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

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

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

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

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

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

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

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

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

\[\boxed{\text{424.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{x}{\sqrt{5}} = \frac{x \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{x\sqrt{5}}{5}\]

\[\textbf{б)}\ \frac{3}{\sqrt{b}} = \frac{3 \cdot \sqrt{b}}{\sqrt{b} \cdot \sqrt{b}} = \frac{3\sqrt{b}}{b}\]

\[\textbf{в)}\ \frac{2}{7\sqrt{y}} = \frac{2 \cdot \sqrt{y}}{7\sqrt{y} \cdot \sqrt{y}} = \frac{2\sqrt{y}}{7y}\]

\[\textbf{г)}\ \frac{a}{b\sqrt{b}} = \frac{a \cdot \sqrt{b}}{b\sqrt{b} \cdot \sqrt{b}} \cdot = \frac{a\sqrt{b}}{b^{2}}\]

\[\textbf{д)}\ \frac{4}{\sqrt{a + b}} = \frac{4 \cdot \sqrt{a + b}}{\sqrt{a + b} \cdot \sqrt{a + b}} =\]

\[= \frac{4\sqrt{a + b}}{a + b}\]

\[\textbf{е)}\ \frac{1}{\sqrt{a - b}} = \frac{1 \cdot \sqrt{a - b}}{\sqrt{a - b} \cdot \sqrt{a - b}} =\]

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

\[\textbf{ж)}\ \frac{5}{2\sqrt{3}} = \frac{5 \cdot \sqrt{3}}{2\sqrt{3} \cdot \sqrt{3}} = \frac{5\sqrt{3}}{2 \cdot 3} =\]

\[= \frac{5\sqrt{3}}{6}\]

\[\textbf{з)}\ \frac{8}{3\sqrt{2}} = \frac{8 \cdot \sqrt{2}}{3\sqrt{2} \cdot \sqrt{2}} = \frac{8\sqrt{2}}{3 \cdot 2} =\]

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

\[\textbf{и)}\ \frac{3\sqrt{5}}{5\sqrt{2}} = \frac{3\sqrt{5} \cdot \sqrt{2}}{5\sqrt{2} \cdot \sqrt{2}} = \frac{3\sqrt{10}}{5 \cdot 2} =\]

\[= \frac{3\sqrt{10}}{10}\]