Страница 20 номер 1.3.14, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Вычислить, используя свойство корня: 1) \(\sqrt{\frac{25}{36}}\); 2) \(\sqrt{\frac{9}{64}}\); 3) \(\sqrt{1\frac{9}{16}}\); 4) \(\sqrt{3\frac{6}{25}}\); 5) \(\sqrt{9 \cdot 4} - \sqrt{\frac{9}{4}}\); 6) \(\sqrt{\frac{16}{25}} - \sqrt{16 \cdot 25}\); 7) \(\frac{\sqrt{32}}{\sqrt{2}}\); 8) \(\frac{\sqrt{5}}{\sqrt{45}}\); 9) \(\frac{\sqrt{(-7)^4}}{\sqrt{(-7)^2}}\); 10) \(\frac{\sqrt{(-15)^6}}{\sqrt{(-15)^2}}\).
1) \(\sqrt{\frac{25}{36}} = \frac{\sqrt{25}}{\sqrt{36}} = \frac{5}{6}\);
2) \(\sqrt{\frac{9}{64}} = \frac{\sqrt{9}}{\sqrt{64}} = \frac{3}{8}\);
3) \(\sqrt{1\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4} = 1\frac{1}{4}\);
4) \(\sqrt{3\frac{6}{25}} = \sqrt{\frac{81}{25}} = \frac{\sqrt{81}}{\sqrt{25}} = \frac{9}{5} = 1\frac{4}{5}\);
5) \(\sqrt{9 \cdot 4} - \sqrt{\frac{9}{4}} = \sqrt{9} \cdot \sqrt{4} - \frac{\sqrt{9}}{\sqrt{4}} = 6 - \frac{3}{2} = 4\frac{1}{2}\);
6) \(\sqrt{\frac{16}{25}} - \sqrt{16 \cdot 25} = \frac{\sqrt{16}}{\sqrt{25}} - \sqrt{16} \cdot \sqrt{25} = \frac{4}{5} - 20 = -19\frac{1}{5}\);
7) \(\frac{\sqrt{32}}{\sqrt{2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4\);
8) \(\frac{\sqrt{5}}{\sqrt{45}} = \sqrt{\frac{5}{45}} = \sqrt{\frac{1}{9}} = \frac{1}{3}\);
9) \(\frac{\sqrt{(-7)^4}}{\sqrt{(-7)^2}} = \sqrt{\frac{(-7)^4}{(-7)^2}} = \sqrt{(-7)^2} = \sqrt{49} = 7\);
10) \(\frac{\sqrt{(-15)^6}}{\sqrt{(-15)^2}} = \sqrt{\frac{(-15)^6}{(-15)^2}} = \sqrt{(-15)^4} = \sqrt{225^2} = 225\).
Ответ: 1) \(\frac{5}{6}\); 2) \(\frac{3}{8}\); 3) \(1\frac{1}{4}\); 4) \(1\frac{4}{5}\); 5) \(4\frac{1}{2}\); 6) \(-19\frac{1}{5}\); 7) \(4\); 8) \(\frac{1}{3}\); 9) \(7\); 10) \(225\).
Вычислить, используя свойство корня: 1) \(\sqrt{\frac{25}{36}}\); 2) \(\sqrt{\frac{9}{64}}\); 3) \(\sqrt{1\frac{9}{16}}\); 4) \(\sqrt{3\frac{6}{25}}\); 5) \(\sqrt{9 \cdot 4} - \sqrt{\frac{9}{4}}\); 6) \(\sqrt{\frac{16}{25}} - \sqrt{16 \cdot 25}\); 7) \(\frac{\sqrt{32}}{\sqrt{2}}\); 8) \(\frac{\sqrt{5}}{\sqrt{45}}\); 9) \(\frac{\sqrt{(-7)^4}}{\sqrt{(-7)^2}}\); 10) \(\frac{\sqrt{(-15)^6}}{\sqrt{(-15)^2}}\).
Пользуемся свойствами арифметического квадратного корня: \(\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}\) при \(a \geqslant 0\), \(b \geqslant 0\); \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\) при \(a \geqslant 0\), \(b > 0\).
1) \(\sqrt{\frac{25}{36}} = \frac{\sqrt{25}}{\sqrt{36}} = \frac{5}{6}\);
2) \(\sqrt{\frac{9}{64}} = \frac{\sqrt{9}}{\sqrt{64}} = \frac{3}{8}\);
3) \(\sqrt{1\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4} = 1\frac{1}{4}\);
4) \(\sqrt{3\frac{6}{25}} = \sqrt{\frac{81}{25}} = \frac{\sqrt{81}}{\sqrt{25}} = \frac{9}{5} = 1\frac{4}{5}\);
5) \(\sqrt{9 \cdot 4} - \sqrt{\frac{9}{4}} = \sqrt{9} \cdot \sqrt{4} - \frac{\sqrt{9}}{\sqrt{4}} = 6 - \frac{3}{2} = 4\frac{1}{2}\);
6) \(\sqrt{\frac{16}{25}} - \sqrt{16 \cdot 25} = \frac{\sqrt{16}}{\sqrt{25}} - \sqrt{16} \cdot \sqrt{25} = \frac{4}{5} - 20 = -19\frac{1}{5}\);
7) \(\frac{\sqrt{32}}{\sqrt{2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4\);
8) \(\frac{\sqrt{5}}{\sqrt{45}} = \sqrt{\frac{5}{45}} = \sqrt{\frac{1}{9}} = \frac{1}{3}\);
9) \(\frac{\sqrt{(-7)^4}}{\sqrt{(-7)^2}} = \sqrt{\frac{(-7)^4}{(-7)^2}} = \sqrt{(-7)^2} = \sqrt{49} = 7\);
10) \(\frac{\sqrt{(-15)^6}}{\sqrt{(-15)^2}} = \sqrt{\frac{(-15)^6}{(-15)^2}} = \sqrt{(-15)^4} = \sqrt{225^2} = 225\).
Ответ: 1) \(\frac{5}{6}\); 2) \(\frac{3}{8}\); 3) \(1\frac{1}{4}\); 4) \(1\frac{4}{5}\); 5) \(4\frac{1}{2}\); 6) \(-19\frac{1}{5}\); 7) \(4\); 8) \(\frac{1}{3}\); 9) \(7\); 10) \(225\).
