Страница 22 номер 1.3.28, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Избавиться от иррациональности в знаменателе дроби: 1) \(\frac{3}{\sqrt{6} - \sqrt{5}}\); 2) \(\frac{7}{\sqrt{15} + \sqrt{14}}\); 3) \(\frac{\sqrt{8} + \sqrt{2}}{\sqrt{8} - \sqrt{2}}\); 4) \(\frac{\sqrt{27} - \sqrt{3}}{\sqrt{27} + \sqrt{3}}\); 5) \(\frac{\sqrt{10}}{\sqrt{10} - 2}\); 6) \(\frac{\sqrt{7}}{3 + \sqrt{7}}\); 7) \(\frac{\sqrt{19}}{4 - \sqrt{19}}\); 8) \(\frac{\sqrt{29}}{5 - \sqrt{29}}\).
1)
\[\begin{aligned} &\frac{3}{\sqrt{6} - \sqrt{5}} = \frac{3\left(\sqrt{6} + \sqrt{5}\right)}{\left(\sqrt{6} - \sqrt{5}\right)\left(\sqrt{6} + \sqrt{5}\right)} = {} \\ &= \frac{3\left(\sqrt{6} + \sqrt{5}\right)}{6 - 5} = 3\left(\sqrt{6} + \sqrt{5}\right). \end{aligned}\]
2)
\[\begin{aligned} &\frac{7}{\sqrt{15} + \sqrt{14}} = \frac{7\left(\sqrt{15} - \sqrt{14}\right)}{\left(\sqrt{15} + \sqrt{14}\right)\left(\sqrt{15} - \sqrt{14}\right)} = {} \\ &= \frac{7\left(\sqrt{15} - \sqrt{14}\right)}{15 - 14} = 7\left(\sqrt{15} - \sqrt{14}\right). \end{aligned}\]
3)
\[\frac{\sqrt{8} + \sqrt{2}}{\sqrt{8} - \sqrt{2}} = \frac{2\sqrt{2} + \sqrt{2}}{2\sqrt{2} - \sqrt{2}} = \frac{3\sqrt{2}}{\sqrt{2}} = 3.\]
4)
\[\frac{\sqrt{27} - \sqrt{3}}{\sqrt{27} + \sqrt{3}} = \frac{3\sqrt{3} - \sqrt{3}}{3\sqrt{3} + \sqrt{3}} = \frac{2\sqrt{3}}{4\sqrt{3}} = \frac{1}{2}.\]
5)
\[\begin{aligned} &\frac{\sqrt{10}}{\sqrt{10} - 2} = \frac{\sqrt{10}\left(\sqrt{10} + 2\right)}{\left(\sqrt{10} - 2\right)\left(\sqrt{10} + 2\right)} = {} \\ &= \frac{10 + 2\sqrt{10}}{10 - 4} = \frac{2\left(5 + \sqrt{10}\right)}{6} = \frac{5 + \sqrt{10}}{3}. \end{aligned}\]
6)
\[\begin{aligned} &\frac{\sqrt{7}}{3 + \sqrt{7}} = \frac{\sqrt{7}\left(3 - \sqrt{7}\right)}{\left(3 + \sqrt{7}\right)\left(3 - \sqrt{7}\right)} = {} \\ &= \frac{\sqrt{7}\left(3 - \sqrt{7}\right)}{9 - 7} = \frac{\sqrt{7}\left(3 - \sqrt{7}\right)}{2}. \end{aligned}\]
7)
\[\begin{aligned} &\frac{\sqrt{19}}{4 - \sqrt{19}} = \frac{\sqrt{19}\left(4 + \sqrt{19}\right)}{\left(4 - \sqrt{19}\right)\left(4 + \sqrt{19}\right)} = {} \\ &= \frac{\sqrt{19}\left(4 + \sqrt{19}\right)}{16 - 19} = -\frac{\sqrt{19}\left(4 + \sqrt{19}\right)}{3}. \end{aligned}\]
8)
\[\begin{aligned} &\frac{\sqrt{29}}{5 - \sqrt{29}} = \frac{\sqrt{29}\left(5 + \sqrt{29}\right)}{\left(5 - \sqrt{29}\right)\left(5 + \sqrt{29}\right)} = {} \\ &= \frac{\sqrt{29}\left(5 + \sqrt{29}\right)}{25 - 29} = -\frac{\sqrt{29}\left(5 + \sqrt{29}\right)}{4}. \end{aligned}\]
Ответ: 1) \(3\left(\sqrt{6} + \sqrt{5}\right)\); 2) \(7\left(\sqrt{15} - \sqrt{14}\right)\); 3) \(3\); 4) \(\frac{1}{2}\); 5) \(\frac{5 + \sqrt{10}}{3}\); 6) \(\frac{\sqrt{7}\left(3 - \sqrt{7}\right)}{2}\); 7) \(-\frac{\sqrt{19}\left(4 + \sqrt{19}\right)}{3}\); 8) \(-\frac{\sqrt{29}\left(5 + \sqrt{29}\right)}{4}\).
Избавиться от иррациональности в знаменателе дроби: 1) \(\frac{3}{\sqrt{6} - \sqrt{5}}\); 2) \(\frac{7}{\sqrt{15} + \sqrt{14}}\); 3) \(\frac{\sqrt{8} + \sqrt{2}}{\sqrt{8} - \sqrt{2}}\); 4) \(\frac{\sqrt{27} - \sqrt{3}}{\sqrt{27} + \sqrt{3}}\); 5) \(\frac{\sqrt{10}}{\sqrt{10} - 2}\); 6) \(\frac{\sqrt{7}}{3 + \sqrt{7}}\); 7) \(\frac{\sqrt{19}}{4 - \sqrt{19}}\); 8) \(\frac{\sqrt{29}}{5 - \sqrt{29}}\).
Умножение числителя и знаменателя на выражение, сопряжённое знаменателю, обращает знаменатель в разность квадратов: \(\left(\sqrt{a} - \sqrt{b}\right)\left(\sqrt{a} + \sqrt{b}\right) = a - b\).
1)
\[\begin{aligned} &\frac{3}{\sqrt{6} - \sqrt{5}} = \frac{3\left(\sqrt{6} + \sqrt{5}\right)}{\left(\sqrt{6} - \sqrt{5}\right)\left(\sqrt{6} + \sqrt{5}\right)} = {} \\ &= \frac{3\left(\sqrt{6} + \sqrt{5}\right)}{6 - 5} = 3\left(\sqrt{6} + \sqrt{5}\right). \end{aligned}\]
2)
\[\begin{aligned} &\frac{7}{\sqrt{15} + \sqrt{14}} = \frac{7\left(\sqrt{15} - \sqrt{14}\right)}{\left(\sqrt{15} + \sqrt{14}\right)\left(\sqrt{15} - \sqrt{14}\right)} = {} \\ &= \frac{7\left(\sqrt{15} - \sqrt{14}\right)}{15 - 14} = 7\left(\sqrt{15} - \sqrt{14}\right). \end{aligned}\]
3)
\[\frac{\sqrt{8} + \sqrt{2}}{\sqrt{8} - \sqrt{2}} = \frac{2\sqrt{2} + \sqrt{2}}{2\sqrt{2} - \sqrt{2}} = \frac{3\sqrt{2}}{\sqrt{2}} = 3.\]
4)
\[\frac{\sqrt{27} - \sqrt{3}}{\sqrt{27} + \sqrt{3}} = \frac{3\sqrt{3} - \sqrt{3}}{3\sqrt{3} + \sqrt{3}} = \frac{2\sqrt{3}}{4\sqrt{3}} = \frac{1}{2}.\]
5)
\[\begin{aligned} &\frac{\sqrt{10}}{\sqrt{10} - 2} = \frac{\sqrt{10}\left(\sqrt{10} + 2\right)}{\left(\sqrt{10} - 2\right)\left(\sqrt{10} + 2\right)} = {} \\ &= \frac{10 + 2\sqrt{10}}{10 - 4} = \frac{2\left(5 + \sqrt{10}\right)}{6} = \frac{5 + \sqrt{10}}{3}. \end{aligned}\]
6)
\[\begin{aligned} &\frac{\sqrt{7}}{3 + \sqrt{7}} = \frac{\sqrt{7}\left(3 - \sqrt{7}\right)}{\left(3 + \sqrt{7}\right)\left(3 - \sqrt{7}\right)} = {} \\ &= \frac{\sqrt{7}\left(3 - \sqrt{7}\right)}{9 - 7} = \frac{\sqrt{7}\left(3 - \sqrt{7}\right)}{2}. \end{aligned}\]
7)
\[\begin{aligned} &\frac{\sqrt{19}}{4 - \sqrt{19}} = \frac{\sqrt{19}\left(4 + \sqrt{19}\right)}{\left(4 - \sqrt{19}\right)\left(4 + \sqrt{19}\right)} = {} \\ &= \frac{\sqrt{19}\left(4 + \sqrt{19}\right)}{16 - 19} = -\frac{\sqrt{19}\left(4 + \sqrt{19}\right)}{3}. \end{aligned}\]
8)
\[\begin{aligned} &\frac{\sqrt{29}}{5 - \sqrt{29}} = \frac{\sqrt{29}\left(5 + \sqrt{29}\right)}{\left(5 - \sqrt{29}\right)\left(5 + \sqrt{29}\right)} = {} \\ &= \frac{\sqrt{29}\left(5 + \sqrt{29}\right)}{25 - 29} = -\frac{\sqrt{29}\left(5 + \sqrt{29}\right)}{4}. \end{aligned}\]
Ответ: 1) \(3\left(\sqrt{6} + \sqrt{5}\right)\); 2) \(7\left(\sqrt{15} - \sqrt{14}\right)\); 3) \(3\); 4) \(\frac{1}{2}\); 5) \(\frac{5 + \sqrt{10}}{3}\); 6) \(\frac{\sqrt{7}\left(3 - \sqrt{7}\right)}{2}\); 7) \(-\frac{\sqrt{19}\left(4 + \sqrt{19}\right)}{3}\); 8) \(-\frac{\sqrt{29}\left(5 + \sqrt{29}\right)}{4}\).
