Страница 46 номер 94, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением
Вычислить (93—94). 1) \(\sqrt{9 + \sqrt{17}} \cdot \sqrt{9 - \sqrt{17}}\); 2) \(\left(\sqrt{3 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}\right)^2\); 3) \(\left(\sqrt{5 + \sqrt{21}} + \sqrt{5 - \sqrt{21}}\right)^2\); 4) \(\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\).
При решении задач будем использовать свойства квадратного корня и формулы сокращенного умножения
1)
\(\sqrt{9 + \sqrt{17}} \cdot \sqrt{9 - \sqrt{17}}\)
\(\sqrt{9 + \sqrt{17}} \cdot \sqrt{9 - \sqrt{17}} = \sqrt{\left(9 + \sqrt{17}\right)\left(9 - \sqrt{17}\right)} =\)
\(= \sqrt{9^{2} - \left(\sqrt{17}\right)^{2}} = \sqrt{81 - 17} = \sqrt{64} = 8\)
2)
\(\left(\sqrt{3 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}\right)^{2}\)
\(\left(\sqrt{3 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}\right)^{2} =\)
\(= \left(\sqrt{3 + \sqrt{5}}\right)^{2} - 2 \cdot \sqrt{3 + \sqrt{5}} \cdot \sqrt{3 - \sqrt{5}} +\)
\(+\left(\sqrt{3 - \sqrt{5}}\right)^{2} = 3 + \sqrt{5} - 2 \cdot \sqrt{\left(3 + \sqrt{5}\right)\left(3 - \sqrt{5}\right)} + 3 - \sqrt{5} =\)
\(= 6 - 2 \cdot \sqrt{3^{2} - \left(\sqrt{5}\right)^{2}} = 6 - 2 \cdot \sqrt{9 - 5} = 6 - 4 = 2\)
3)
\(\left(\sqrt{5 + \sqrt{21}} + \sqrt{5 - \sqrt{21}}\right)^{2}\)
\(\left(\sqrt{5 + \sqrt{21}} + \sqrt{5 - \sqrt{21}}\right)^{2} = \left(\sqrt{5 + \sqrt{21}}\right)^{2} + 2 \cdot \sqrt{5 + \sqrt{21}} \cdot\)
\(\cdot \sqrt{5 - \sqrt{21}} + \left(\sqrt{5 - \sqrt{21}}\right)^{2} =\)
\(= 5 + \sqrt{21} + 2 \cdot \sqrt{\left(5 + \sqrt{21}\right)\left(5 - \sqrt{21}\right)} +\)
\(+5 - \sqrt{21} = 10 + 2 \cdot \sqrt{5^{2} - \left(\sqrt{21}\right)^{2}} =\)
\(= 10 + 2 \cdot \sqrt{25 - 21} = 10 + 2 \cdot 2 = 14\)
4)
\(\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\)
Приведем наши дроби к общему знаменателю:
\(\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} =\)
\(= \dfrac{\left(\sqrt{3} + \sqrt{2}\right)\left(\sqrt{3} + \sqrt{2}\right) - \left(\sqrt{3} - \sqrt{2}\right)\left(\sqrt{3} - \sqrt{2}\right)}{\left(\sqrt{3} - \sqrt{2}\right)\left(\sqrt{3} + \sqrt{2}\right)} =\)
\(= \dfrac{\left(\sqrt{3} + \sqrt{2}\right)^{2} - \left(\sqrt{3} - \sqrt{2}\right)^{2}}{\left(\sqrt{3}\right)^{2} - \left(\sqrt{2}\right)^{2}} =\)
\(= \dfrac{\left(\left.\sqrt{3} + \sqrt{2}\right. - \left.\left(\sqrt{3} - \sqrt{2}\right)\right.\right) \cdot \left(\sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2}\right)}{3 - 2} =\)
\(= \dfrac{\left(\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2}\right) \cdot 2\sqrt{3}}{1} =\)
\(= \left(\sqrt{2} + \sqrt{2}\right) \cdot 2\sqrt{3} = 2\sqrt{2} \cdot 2\sqrt{3} = 4\sqrt{2 \cdot 3} = 4\sqrt{6}\)
