Страница 20 номер 1.3.15, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Представить в виде дроби с целым знаменателем: 1) \(\frac{1}{\sqrt{7}}\); 2) \(\frac{1}{\sqrt{13}}\); 3) \(-\frac{3}{\sqrt{11}}\); 4) \(-\frac{5}{\sqrt{17}}\); 5) \(\frac{6}{\sqrt{15}}\); 6) \(\frac{6}{\sqrt{21}}\); 7) \(\frac{\sqrt{2}}{\sqrt{6}}\); 8) \(\frac{\sqrt{3}}{\sqrt{15}}\).
Числитель и знаменатель умножаем на корень из знаменателя.
1) \(\frac{1}{\sqrt{7}} = \frac{\sqrt{7}}{\left(\sqrt{7}\right)^2} = \frac{\sqrt{7}}{7}\);
2) \(\frac{1}{\sqrt{13}} = \frac{\sqrt{13}}{\left(\sqrt{13}\right)^2} = \frac{\sqrt{13}}{13}\);
3) \(-\frac{3}{\sqrt{11}} = -\frac{3\sqrt{11}}{\left(\sqrt{11}\right)^2} = -\frac{3\sqrt{11}}{11}\);
4) \(-\frac{5}{\sqrt{17}} = -\frac{5\sqrt{17}}{\left(\sqrt{17}\right)^2} = -\frac{5\sqrt{17}}{17}\);
5) \(\frac{6}{\sqrt{15}} = \frac{6\sqrt{15}}{\left(\sqrt{15}\right)^2} = \frac{6\sqrt{15}}{15} = \frac{2\sqrt{15}}{5}\);
6) \(\frac{6}{\sqrt{21}} = \frac{6\sqrt{21}}{\left(\sqrt{21}\right)^2} = \frac{6\sqrt{21}}{21} = \frac{2\sqrt{21}}{7}\);
7) \(\frac{\sqrt{2}}{\sqrt{6}} = \sqrt{\frac{2}{6}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\);
8) \(\frac{\sqrt{3}}{\sqrt{15}} = \sqrt{\frac{3}{15}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\).
Ответ: 1) \(\frac{\sqrt{7}}{7}\); 2) \(\frac{\sqrt{13}}{13}\); 3) \(-\frac{3\sqrt{11}}{11}\); 4) \(-\frac{5\sqrt{17}}{17}\); 5) \(\frac{2\sqrt{15}}{5}\); 6) \(\frac{2\sqrt{21}}{7}\); 7) \(\frac{\sqrt{3}}{3}\); 8) \(\frac{\sqrt{5}}{5}\).
Представить в виде дроби с целым знаменателем: 1) \(\frac{1}{\sqrt{7}}\); 2) \(\frac{1}{\sqrt{13}}\); 3) \(-\frac{3}{\sqrt{11}}\); 4) \(-\frac{5}{\sqrt{17}}\); 5) \(\frac{6}{\sqrt{15}}\); 6) \(\frac{6}{\sqrt{21}}\); 7) \(\frac{\sqrt{2}}{\sqrt{6}}\); 8) \(\frac{\sqrt{3}}{\sqrt{15}}\).
Чтобы в знаменателе не осталось знака корня, умножаем числитель и знаменатель дроби на корень из знаменателя и пользуемся равенством \(\left(\sqrt{a}\right)^2 = a\).
1) \(\frac{1}{\sqrt{7}} = \frac{\sqrt{7}}{\left(\sqrt{7}\right)^2} = \frac{\sqrt{7}}{7}\);
2) \(\frac{1}{\sqrt{13}} = \frac{\sqrt{13}}{\left(\sqrt{13}\right)^2} = \frac{\sqrt{13}}{13}\);
3) \(-\frac{3}{\sqrt{11}} = -\frac{3\sqrt{11}}{\left(\sqrt{11}\right)^2} = -\frac{3\sqrt{11}}{11}\);
4) \(-\frac{5}{\sqrt{17}} = -\frac{5\sqrt{17}}{\left(\sqrt{17}\right)^2} = -\frac{5\sqrt{17}}{17}\);
5) \(\frac{6}{\sqrt{15}} = \frac{6\sqrt{15}}{\left(\sqrt{15}\right)^2} = \frac{6\sqrt{15}}{15} = \frac{2\sqrt{15}}{5}\);
6) \(\frac{6}{\sqrt{21}} = \frac{6\sqrt{21}}{\left(\sqrt{21}\right)^2} = \frac{6\sqrt{21}}{21} = \frac{2\sqrt{21}}{7}\);
7) \(\frac{\sqrt{2}}{\sqrt{6}} = \sqrt{\frac{2}{6}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\);
8) \(\frac{\sqrt{3}}{\sqrt{15}} = \sqrt{\frac{3}{15}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\).
Ответ: 1) \(\frac{\sqrt{7}}{7}\); 2) \(\frac{\sqrt{13}}{13}\); 3) \(-\frac{3\sqrt{11}}{11}\); 4) \(-\frac{5\sqrt{17}}{17}\); 5) \(\frac{2\sqrt{15}}{5}\); 6) \(\frac{2\sqrt{21}}{7}\); 7) \(\frac{\sqrt{3}}{3}\); 8) \(\frac{\sqrt{5}}{5}\).
