Question

multumesc mult de tot ​

Answer 1

Explicație pas cu pas:

$\sqrt{ {\Big( \dfrac{1}{125}\Big)}^{ \frac{5}{2} } } = \sqrt{ {\Big( \dfrac{1}{ {5}^{3} }\Big)}^{ \frac{5}{2} } } = \sqrt{ {\Big( {5}^{ - 3} \Big)}^{ \frac{5}{2} } } = \sqrt{ {5}^{ - \frac{15}{2} } } = {5}^{ - \frac{15}{4} }$

$\sqrt[3]{{\Big( \dfrac{1}{25} \Big)}^{5}} = \sqrt[3]{{\Big( \dfrac{1}{ {5}^{2} } \Big)}^{5}} = \sqrt[3]{{\Big({5}^{ - 2} \Big)}^{5}} = \sqrt[3]{{5}^{ - 10}} = {5}^{ - \frac{10}{3} }$

avem aceeași bază (5) și comparăm puterile (le aducem la același numitor):

$- \dfrac{^{3)} 15}{4} = - \dfrac{45}{12}$

$- \dfrac{^{4)}10}{3} = - \dfrac{40}{12}$

=>

$- 45 < - 40 \implies - \dfrac{45}{12} < - \dfrac{40}{12} \\ \implies - \dfrac{15}{4} < - \dfrac{10}{3} \implies {5}^{ - \frac{15}{4} } < {5}^{ - \frac{10}{3} }$

$\implies \sqrt{ {\Big( \dfrac{1}{125}\Big)}^{ \frac{5}{2} } } < \sqrt[3]{{\Big( \dfrac{1}{25} \Big)}^{5}}$