Question

Se pricepe cineva la mate sa mi explice și ie punctele b

Answer 1

Explicație pas cu pas:

${\bigg( \sqrt{ \dfrac{27}{25} } \bigg)}^{- 5} = {\bigg( \sqrt{ \dfrac{25}{27} } \bigg)}^{5} = {\bigg( \sqrt{ \dfrac{ {5}^{2} }{ {3}^{3} } } \bigg)}^{5} = \sqrt{{\bigg(\dfrac{ {5}^{2} }{ {3}^{3} } \bigg)}^{5}} = \sqrt{ \dfrac{ {( {5}^{2} )}^{5} }{ {( {3}^{3} )}^{5} } } = \sqrt{ \dfrac{{5}^{10}}{ {3}^{15} }}$

$\dfrac{ {( \sqrt{5} )}^{3} }{ {( \sqrt{3} )}^{8} } = \dfrac{ \sqrt{{5}^{3}} }{ \sqrt{{3}^{8}} } = \sqrt{ \dfrac{ {5}^{3} }{ {3}^{8} } }$

=>

${\bigg( \sqrt{ \dfrac{27}{25} } \bigg)}^{- 5} : \dfrac{ {( \sqrt{5} )}^{3} }{ {( \sqrt{3} )}^{8} } = \sqrt{ \dfrac{{5}^{10}}{ {3}^{15} }} : \sqrt{ \dfrac{ {5}^{3} }{ {3}^{8} } } = \sqrt{ \dfrac{{5}^{10}}{ {3}^{15} }} \cdot \sqrt{ \dfrac{ {3}^{8} }{ {5}^{3} } } = \sqrt{ \dfrac{{5}^{10}}{ {3}^{15} } \cdot \dfrac{ {3}^{8} }{ {5}^{3} }} = \sqrt{ \dfrac{{5}^{10 - 3}}{ {3}^{15 - 8} }} = \sqrt{ \dfrac{{5}^{7}}{ {3}^{7} }} = \sqrt{ {\bigg( \dfrac{5}{3} \bigg)}^{7} } = {\bigg( \sqrt{\dfrac{5}{3}} \bigg)}^{7}$