Страница 19 номер 1.3.11, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Внести множитель под знак корня: 1) \(2\sqrt{5}\); 2) \(3\sqrt{7}\); 3) \(5\sqrt{3}\); 4) \(4\sqrt{2}\); 5) \(0{,}1\sqrt{6}\); 6) \(0{,}2\sqrt{3}\); 7) \(1{,}2\sqrt{2}\); 8) \(1{,}1\sqrt{5}\); 9) \(\frac{1}{2}\sqrt{6}\); 10) \(\frac{2}{3}\sqrt{18}\); 11) \(1\frac{1}{7}\sqrt{\frac{7}{16}}\); 12) \(1\frac{2}{5}\sqrt{\frac{10}{21}}\).
1) \(2\sqrt{5} = \sqrt{4} \cdot \sqrt{5} = \sqrt{20}\);
2) \(3\sqrt{7} = \sqrt{9} \cdot \sqrt{7} = \sqrt{63}\);
3) \(5\sqrt{3} = \sqrt{25} \cdot \sqrt{3} = \sqrt{75}\);
4) \(4\sqrt{2} = \sqrt{16} \cdot \sqrt{2} = \sqrt{32}\);
5) \(0{,}1\sqrt{6} = \sqrt{0{,}01} \cdot \sqrt{6} = \sqrt{0{,}06}\);
6) \(0{,}2\sqrt{3} = \sqrt{0{,}04} \cdot \sqrt{3} = \sqrt{0{,}12}\);
7) \(1{,}2\sqrt{2} = \sqrt{1{,}44} \cdot \sqrt{2} = \sqrt{2{,}88}\);
8) \(1{,}1\sqrt{5} = \sqrt{1{,}21} \cdot \sqrt{5} = \sqrt{6{,}05}\);
9) \(\frac{1}{2}\sqrt{6} = \sqrt{\frac{1}{4}} \cdot \sqrt{6} = \sqrt{\frac{6}{4}} = \sqrt{\frac{3}{2}}\);
10) \(\frac{2}{3}\sqrt{18} = \sqrt{\frac{4}{9}} \cdot \sqrt{18} = \sqrt{\frac{4 \cdot 18}{9}} = \sqrt{8}\);
11) \(1\frac{1}{7}\sqrt{\frac{7}{16}} = \frac{8}{7}\sqrt{\frac{7}{16}} = \sqrt{\frac{64}{49}} \cdot \sqrt{\frac{7}{16}} = \sqrt{\frac{64 \cdot 7}{49 \cdot 16}} = \sqrt{\frac{4}{7}}\);
12) \(1\frac{2}{5}\sqrt{\frac{10}{21}} = \frac{7}{5}\sqrt{\frac{10}{21}} = \sqrt{\frac{49}{25}} \cdot \sqrt{\frac{10}{21}} = \sqrt{\frac{49 \cdot 10}{25 \cdot 21}} = \sqrt{\frac{14}{15}}\).
Ответ: 1) \(\sqrt{20}\); 2) \(\sqrt{63}\); 3) \(\sqrt{75}\); 4) \(\sqrt{32}\); 5) \(\sqrt{0{,}06}\); 6) \(\sqrt{0{,}12}\); 7) \(\sqrt{2{,}88}\); 8) \(\sqrt{6{,}05}\); 9) \(\sqrt{\frac{3}{2}}\); 10) \(\sqrt{8}\); 11) \(\sqrt{\frac{4}{7}}\); 12) \(\sqrt{\frac{14}{15}}\).
Внести множитель под знак корня: 1) \(2\sqrt{5}\); 2) \(3\sqrt{7}\); 3) \(5\sqrt{3}\); 4) \(4\sqrt{2}\); 5) \(0{,}1\sqrt{6}\); 6) \(0{,}2\sqrt{3}\); 7) \(1{,}2\sqrt{2}\); 8) \(1{,}1\sqrt{5}\); 9) \(\frac{1}{2}\sqrt{6}\); 10) \(\frac{2}{3}\sqrt{18}\); 11) \(1\frac{1}{7}\sqrt{\frac{7}{16}}\); 12) \(1\frac{2}{5}\sqrt{\frac{10}{21}}\).
Положительный множитель, стоящий перед корнем, заменяем равным ему арифметическим квадратным корнем из его квадрата и перемножаем корни: \(b\sqrt{a} = \sqrt{b^2} \cdot \sqrt{a} = \sqrt{b^2 a}\).
1) \(2\sqrt{5} = \sqrt{4} \cdot \sqrt{5} = \sqrt{20}\);
2) \(3\sqrt{7} = \sqrt{9} \cdot \sqrt{7} = \sqrt{63}\);
3) \(5\sqrt{3} = \sqrt{25} \cdot \sqrt{3} = \sqrt{75}\);
4) \(4\sqrt{2} = \sqrt{16} \cdot \sqrt{2} = \sqrt{32}\);
5) \(0{,}1\sqrt{6} = \sqrt{0{,}01} \cdot \sqrt{6} = \sqrt{0{,}06}\);
6) \(0{,}2\sqrt{3} = \sqrt{0{,}04} \cdot \sqrt{3} = \sqrt{0{,}12}\);
7) \(1{,}2\sqrt{2} = \sqrt{1{,}44} \cdot \sqrt{2} = \sqrt{2{,}88}\);
8) \(1{,}1\sqrt{5} = \sqrt{1{,}21} \cdot \sqrt{5} = \sqrt{6{,}05}\);
9) \(\frac{1}{2}\sqrt{6} = \sqrt{\frac{1}{4}} \cdot \sqrt{6} = \sqrt{\frac{6}{4}} = \sqrt{\frac{3}{2}}\);
10) \(\frac{2}{3}\sqrt{18} = \sqrt{\frac{4}{9}} \cdot \sqrt{18} = \sqrt{\frac{4 \cdot 18}{9}} = \sqrt{8}\);
11) \(1\frac{1}{7}\sqrt{\frac{7}{16}} = \frac{8}{7}\sqrt{\frac{7}{16}} = \sqrt{\frac{64}{49}} \cdot \sqrt{\frac{7}{16}} = \sqrt{\frac{64 \cdot 7}{49 \cdot 16}} = \sqrt{\frac{4}{7}}\);
12) \(1\frac{2}{5}\sqrt{\frac{10}{21}} = \frac{7}{5}\sqrt{\frac{10}{21}} = \sqrt{\frac{49}{25}} \cdot \sqrt{\frac{10}{21}} = \sqrt{\frac{49 \cdot 10}{25 \cdot 21}} = \sqrt{\frac{14}{15}}\).
Ответ: 1) \(\sqrt{20}\); 2) \(\sqrt{63}\); 3) \(\sqrt{75}\); 4) \(\sqrt{32}\); 5) \(\sqrt{0{,}06}\); 6) \(\sqrt{0{,}12}\); 7) \(\sqrt{2{,}88}\); 8) \(\sqrt{6{,}05}\); 9) \(\sqrt{\frac{3}{2}}\); 10) \(\sqrt{8}\); 11) \(\sqrt{\frac{4}{7}}\); 12) \(\sqrt{\frac{14}{15}}\).
