Решебник по алгебре 8 класс Макарычев ФГОС Задание 370

 

\[\boxed{\text{370\ (370).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \sqrt{\frac{9}{64}} = \frac{\sqrt{9}}{\sqrt{64}} = \frac{3}{8}\]

\[\textbf{б)}\ \sqrt{\frac{36}{25}} = \frac{\sqrt{36}}{\sqrt{25}} = \frac{6}{5} = 1,2\]

\[\textbf{в)}\ \sqrt{\frac{121}{25}} = \frac{\sqrt{121}}{\sqrt{25}} = \frac{11}{5} = 2,2\]

\[\textbf{г)}\ \sqrt{1\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4} =\]

\[= 1\frac{1}{4} = 1,25\]

\[\textbf{д)}\ \sqrt{2\frac{7}{81}} = \sqrt{\frac{169}{81}} = \frac{\sqrt{169}}{\sqrt{81}} =\]

\[= \frac{13}{9} = 1\frac{4}{9}\]

\[\textbf{е)}\ \sqrt{5\frac{1}{16}} = \sqrt{\frac{81}{16}} = \frac{\sqrt{81}}{\sqrt{16}} = \frac{9}{4} =\]

\[= 2\frac{1}{4} = 2,25\]

\[\boxed{\text{370.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Если \(a \geq 0\ и\ b \geq 0,\ \)то:

\[\sqrt{\text{ab}} = \sqrt{a} \cdot \sqrt{b}.\]

Если \(a \geq 0\ и\ b > 0\), то:

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

Формула разности квадратов:

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

Решение.

\[\textbf{а)}\ \sqrt{17^{2} - 8^{2}} =\]

\[= \sqrt{(17 - 8)(17 + 8)} =\]

\[= \sqrt{9 \cdot 25} = 3 \cdot 5 = 15\]

\[\textbf{б)}\ \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} =\]

\[= 5\]

\[\textbf{в)}\ \sqrt{82^{2} - 18^{2}} =\]

\[= \sqrt{(82 - 18)(82 + 18)} =\]

\[= \sqrt{64 \cdot 100} = 8 \cdot 10 = 80\]

\[\textbf{г)}\ \sqrt{117^{2} - 108^{2}} =\]

\[= \sqrt{(117 - 108)(117 + 108)} =\]

\[= \sqrt{9 \cdot 225} = 3 \cdot 15 = 45\]

\[\textbf{д)}\ \sqrt{{6,8}^{2} - {3,2}^{2}} =\]

\[= \sqrt{(6,8 - 3,2)(6,8 + 3,2)} =\]

\[= \sqrt{3,6 \cdot 10} = \sqrt{36} = 6\]

\[\textbf{е)}\ \sqrt{\left( 1\frac{1}{16} \right)^{2} - \left( \frac{1}{2} \right)^{2}} =\]

\[= \sqrt{\left( \frac{17}{6} - \frac{1}{2} \right) \cdot \left( \frac{17}{6} + \frac{1}{2} \right)} =\]

\[= \sqrt{\frac{17 - 8}{16} \cdot \frac{17 + 8}{16}} =\]

\[= \sqrt{\frac{9 \cdot 25}{16 \cdot 16}} = \frac{3 \cdot 5}{16} = \frac{15}{16}\]