Страница 45 номер 93, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением
Вычислить (93—94). 1) \(\sqrt[3]{-125} + \frac{1}{8}\sqrt[6]{64}\); 2) \(\sqrt[5]{32} - 0{,}5\sqrt[3]{-216}\); 3) \(-\frac{1}{3}\sqrt[4]{81} + \sqrt[4]{625}\); 4) \(\sqrt[3]{-1000} - \frac{1}{4}\sqrt[4]{256}\); 5) \(\sqrt[4]{0{,}0001} - 2\sqrt{0{,}25} + \sqrt[5]{-\frac{1}{32}}\); 6) \(\sqrt[5]{\frac{1}{243}} + \sqrt[3]{-0{,}001} - \sqrt[4]{0{,}0016}\).
1)
\(\sqrt[3]{-125} + \dfrac{1}{8}\sqrt[6]{64} = \sqrt[3]{(-5)^{3}} + \dfrac{1}{8}\sqrt[6]{2^{6}} = -5 + \dfrac{1}{8} \cdot 2 =\)
\(= -5 + \dfrac{2}{8} = -5 + \dfrac{1}{4} = -\dfrac{20}{4} + \dfrac{1}{4} = -\dfrac{19}{4}\)
2)
\(\sqrt[5]{32} - 0{,}5\sqrt[3]{-216} = \sqrt[5]{2^{5}} - 0{,}5\sqrt[3]{(-6)^{3}} = 2 - \dfrac{1}{2} \cdot \left(-6\right) = 2 + 3 = 5\)
3)
\(-\dfrac{1}{3}\sqrt[4]{81} + \sqrt[4]{625} = -\dfrac{1}{3}\sqrt[4]{3^{4}} + \sqrt[4]{25^{2}} = -\dfrac{1}{3} \cdot 3 + \sqrt[4]{5^{4}} = -1 + 5 = 4\)
4)
\(\sqrt[3]{-1000} - \dfrac{1}{4}\sqrt[4]{256} = \sqrt[3]{(-10)^{3}} - \dfrac{1}{4}\sqrt[4]{4^{4}} =\)
\(= -10 - \dfrac{1}{4} \cdot 4 = -10 - 1 = -11\)
5)
\(\sqrt[4]{0{,}0001} - 2\sqrt{0{,}25} + \sqrt[5]{-\dfrac{1}{32}} = \sqrt[4]{0{,}1^{4}} - 2 \cdot \sqrt{0{,}5^{2}} + \sqrt[5]{\left(\dfrac{1}{-2}\right)^{5}} =\)
\(= 0{,}1 - 2 \cdot \dfrac{1}{2} + \dfrac{1}{-2} = 0{,}1 - 1 - \dfrac{1}{2} = 0{,}1 - 1 - 0{,}5 = -1{,}4\)
6)
\(\sqrt[5]{\dfrac{1}{243}} + \sqrt[3]{-0{,}001} - \sqrt[4]{0{,}0016} = \sqrt[5]{\dfrac{1}{3^{5}}} + \sqrt[3]{(-0{,}1)^{3}} - \sqrt[4]{0{,}2^{4}} =\)
\(= \dfrac{1}{3} + \left(-0{,}1\right) - 0{,}2 = \dfrac{1}{3} - 0{,}3 = \dfrac{1}{3} - \dfrac{3}{10} = \dfrac{10 - 9}{30} = \dfrac{1}{30}\)
