Упр.22.4 ГДЗ Мерзляк Поляков 9 класс (Алгебра)

1) 25^(1/3)·5^(1/3); 4) (9-v17)^(1/3)·(9+v17)^(1/3);
2) (80^(1/4))/5^(1/4); 5) (2v17+10)^(1/5)·(2v17-10)^(1/5);
3) (2^15·5^3)^(1/7)·(2^6·5^4)^(1/7); 6) 125^(1/4)·v18·v2·80^(1/4)?

1. $$\sqrt[3]{25}\cdot \sqrt[3]{5}=\sqrt[3]{25\cdot 5}=\sqrt[3]{125}=\sqrt[3]{5^3}=5.$$
2. $$\frac{\sqrt[4]{80}}{\sqrt[4]{5}}=\sqrt[4]{\frac{80}{5}}=\sqrt[4]{16}=\sqrt[4]{2^4}=2.$$
3. $$ \sqrt[7]{2^{15}\cdot 5^3}\cdot \sqrt[7]{2^6\cdot 5^4} =\sqrt[7]{2^{15}\cdot 5^3\cdot 2^6\cdot 5^4} =\sqrt[7]{2^{21}\cdot 5^7} =\sqrt[7]{(2^3)^7\cdot 5^7} =2^3\cdot 5=40. $$
4. $$ \sqrt[3]{9-\sqrt{17}}\cdot \sqrt[3]{9+\sqrt{17}} =\sqrt[3]{(9-\sqrt{17})(9+\sqrt{17})} =\sqrt[3]{81-17} =\sqrt[3]{64} =4. $$
5. $$ \sqrt[5]{2\sqrt{17}+10}\cdot \sqrt[5]{2\sqrt{17}-10} =\sqrt[5]{(2\sqrt{17}+10)(2\sqrt{17}-10)} =\sqrt[5]{(2\sqrt{17})^2-10^2} =\sqrt[5]{68-100} =\sqrt[5]{-32} =-2. $$
6. $$ \sqrt[4]{125}\cdot \sqrt{18}\cdot \sqrt{2}\cdot \sqrt[4]{80} =\sqrt[4]{125\cdot 80}\cdot \sqrt{18\cdot 2} =\sqrt[4]{5^3\cdot 5\cdot 16}\cdot \sqrt{36} =\sqrt[4]{5^4\cdot 2^4}\cdot 6 =5\cdot 2\cdot 6=60. $$

Ответ

1) 5; 2) 2; 3) 40; 4) 4; 5) −2; 6) 60.