Question

am nevoie de E9 d si e.50 PUNCTE!!!!!

Answer 1

Explicație pas cu pas:

d)

$\sqrt[4]{ \sqrt[6]{32} } = \sqrt[4]{ \sqrt[6]{ {2}^{5} } } = {2}^{ \frac{5}{6 \cdot 4} } = {2}^{ \frac{25}{120} }$

$\sqrt[5]{ \sqrt[6]{16} } = \sqrt[5]{ \sqrt[6]{ {2}^{4} } } = {2}^{ \frac{4}{5 \cdot 6} } = {2}^{ \frac{16}{120} }$

$\frac{25}{120} > \frac{16}{120} \\ \implies \sqrt[4]{ \sqrt[6]{32} } > \sqrt[5]{ \sqrt[6]{16} }$

e)

$\sqrt[3]{23} = 23^{\frac{1}{3}}$

$\sqrt[4]{4} + \sqrt{2} = \sqrt[4]{ {2}^{2} } + \sqrt{2} = {2}^{ \frac{2}{4} } + \sqrt{2} = {2}^{ \frac{1}{2} } + \sqrt{2} = 2\sqrt{2} = \sqrt{2^{3}} = 2^{\frac{3}{2}}$

$\Big(23^{\frac{1}{3}}\Big)^{6} = 23^{\frac{6}{3}} = 23^{2} = 529\\ \Big(2^{\frac{3}{2}}\Big)^{6} = 2^{\frac{18}{2}} = 2^{9} = 512\\ \implies \sqrt[3]{23} > \sqrt[4]{4} + 2$