Упр.880 ГДЗ Никольский Потапов 9 класс (Алгебра)

а) v(v5)·((v5)^(1/3):v(5^(1/4)))^2;
б) ((16(8v2)^(1/4))^(1/3))^2·(32(2^(1/4))^(1/3)·v(2(4^(1/4))(3^(1/3))).

а)

$$ \sqrt{5}\cdot\left(\sqrt[3]{\sqrt{5}}:\sqrt[4]{\sqrt{5}}\right)^2 = 5^{1/2}\cdot\left(5^{1/6}:5^{1/8}\right)^2 $$

$$ =5^{1/2}\cdot\left(5^{1/6-1/8}\right)^2 =5^{1/2}\cdot\left(5^{1/24}\right)^2 =5^{1/2}\cdot 5^{1/12} =5^{7/12} $$

$$ 5^{7/12}=\sqrt[12]{5^7} $$

б)

$$ \left(\sqrt[3]{16\sqrt[4]{8\sqrt{2}}}\right)^2\cdot \sqrt[3]{32\sqrt[4]{2}}\cdot \sqrt{2\sqrt[4]{4}\cdot \sqrt[3]{3}} $$

$$ 16\sqrt[4]{8\sqrt{2}}=2^4\cdot (2^3\cdot 2^{1/2})^{1/4} =2^4\cdot 2^{7/8}=2^{39/8} $$

$$ \left(\sqrt[3]{16\sqrt[4]{8\sqrt{2}}}\right)^2 =\left(2^{39/8}\right)^{2/3}=2^{13/4} $$

$$ \sqrt[3]{32\sqrt[4]{2}}=\left(2^5\cdot 2^{1/4}\right)^{1/3}=2^{7/4} $$

$$ \sqrt{2\sqrt[4]{4}\cdot \sqrt[3]{3}} =\sqrt{2\cdot 2^{1/2}\cdot 3^{1/3}} =2^{3/4}\cdot 3^{1/6} $$

$$ 2^{13/4}\cdot 2^{7/4}\cdot 2^{3/4}\cdot 3^{1/6} =2^{23/4}\cdot 3^{1/6} =32\cdot \sqrt[4]{8}\cdot \sqrt[6]{3} $$

Ответ

а) $$\sqrt[12]{5^7}$$; б) $$32\cdot \sqrt[4]{8}\cdot \sqrt[6]{3}$$.