Страница 148 номер 6.3.2, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
\(q = \frac{b_2}{b_1}\); зная \(q\), находим \(b_3 = b_2 q\) и \(b_4 = b_3 q\).
1) \(q = \frac{5}{3}\); \(b_3 = 5 \cdot \frac{5}{3} = \frac{25}{3} = 8\frac{1}{3}\), \(b_4 = \frac{25}{3} \cdot \frac{5}{3} = \frac{125}{9} = 13\frac{8}{9}\).
2) \(q = \frac{6}{5} = 1{,}2\); \(b_3 = 6 \cdot 1{,}2 = 7{,}2\), \(b_4 = 7{,}2 \cdot 1{,}2 = 8{,}64\).
3) \(q = \frac{2}{6} = \frac{1}{3}\); \(b_3 = 2 \cdot \frac{1}{3} = \frac{2}{3}\), \(b_4 = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}\).
4) \(q = \frac{5}{10} = 0{,}5\); \(b_3 = 5 \cdot 0{,}5 = 2{,}5\), \(b_4 = 2{,}5 \cdot 0{,}5 = 1{,}25\).
5) \(q = \frac{4}{-8} = -\frac{1}{2}\); \(b_3 = 4 \cdot \left(-\frac{1}{2}\right) = -2\), \(b_4 = -2 \cdot \left(-\frac{1}{2}\right) = 1\).
6) \(q = \frac{2}{-6} = -\frac{1}{3}\); \(b_3 = 2 \cdot \left(-\frac{1}{3}\right) = -\frac{2}{3}\), \(b_4 = -\frac{2}{3} \cdot \left(-\frac{1}{3}\right) = \frac{2}{9}\).
7) \(q = -1 : \left(-\frac{1}{3}\right) = 3\); \(b_3 = -1 \cdot 3 = -3\), \(b_4 = -3 \cdot 3 = -9\).
8) В сборнике второй член пункта 8) напечатан как \(b_1\); решаем при \(b_2 = -2\). Тогда \(q = -2 : \left(-\frac{2}{3}\right) = 3\); \(b_3 = -2 \cdot 3 = -6\), \(b_4 = -6 \cdot 3 = -18\).
Ответ: 1) \(b_3 = 8\frac{1}{3}\), \(b_4 = 13\frac{8}{9}\); 2) \(b_3 = 7{,}2\), \(b_4 = 8{,}64\); 3) \(b_3 = \frac{2}{3}\), \(b_4 = \frac{2}{9}\); 4) \(b_3 = 2{,}5\), \(b_4 = 1{,}25\); 5) \(b_3 = -2\), \(b_4 = 1\); 6) \(b_3 = -\frac{2}{3}\), \(b_4 = \frac{2}{9}\); 7) \(b_3 = -3\), \(b_4 = -9\); 8) \(b_3 = -6\), \(b_4 = -18\).
Знаменатель геометрической прогрессии равен отношению любого её члена, начиная со второго, к предыдущему, поэтому \(q = \frac{b_2}{b_1}\). Зная \(q\), находим \(b_3 = b_2 q\) и \(b_4 = b_3 q\).
1) \(q = \frac{5}{3}\); \(b_3 = 5 \cdot \frac{5}{3} = \frac{25}{3} = 8\frac{1}{3}\), \(b_4 = \frac{25}{3} \cdot \frac{5}{3} = \frac{125}{9} = 13\frac{8}{9}\).
2) \(q = \frac{6}{5} = 1{,}2\); \(b_3 = 6 \cdot 1{,}2 = 7{,}2\), \(b_4 = 7{,}2 \cdot 1{,}2 = 8{,}64\).
3) \(q = \frac{2}{6} = \frac{1}{3}\); \(b_3 = 2 \cdot \frac{1}{3} = \frac{2}{3}\), \(b_4 = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}\).
4) \(q = \frac{5}{10} = 0{,}5\); \(b_3 = 5 \cdot 0{,}5 = 2{,}5\), \(b_4 = 2{,}5 \cdot 0{,}5 = 1{,}25\).
5) \(q = \frac{4}{-8} = -\frac{1}{2}\); \(b_3 = 4 \cdot \left(-\frac{1}{2}\right) = -2\), \(b_4 = -2 \cdot \left(-\frac{1}{2}\right) = 1\).
6) \(q = \frac{2}{-6} = -\frac{1}{3}\); \(b_3 = 2 \cdot \left(-\frac{1}{3}\right) = -\frac{2}{3}\), \(b_4 = -\frac{2}{3} \cdot \left(-\frac{1}{3}\right) = \frac{2}{9}\).
7) \(q = -1 : \left(-\frac{1}{3}\right) = 3\); \(b_3 = -1 \cdot 3 = -3\), \(b_4 = -3 \cdot 3 = -9\).
8) В сборнике второй член пункта 8) напечатан как \(b_1\); решаем при \(b_2 = -2\). Тогда \(q = -2 : \left(-\frac{2}{3}\right) = 3\); \(b_3 = -2 \cdot 3 = -6\), \(b_4 = -6 \cdot 3 = -18\).
Ответ: 1) \(b_3 = 8\frac{1}{3}\), \(b_4 = 13\frac{8}{9}\); 2) \(b_3 = 7{,}2\), \(b_4 = 8{,}64\); 3) \(b_3 = \frac{2}{3}\), \(b_4 = \frac{2}{9}\); 4) \(b_3 = 2{,}5\), \(b_4 = 1{,}25\); 5) \(b_3 = -2\), \(b_4 = 1\); 6) \(b_3 = -\frac{2}{3}\), \(b_4 = \frac{2}{9}\); 7) \(b_3 = -3\), \(b_4 = -9\); 8) \(b_3 = -6\), \(b_4 = -18\).
