Răspuns:
[tex]\boldsymbol{ \red{e) \ \dfrac{2}{5}; \ f) \ \dfrac{32}{243}}}[/tex]
Explicație pas cu pas:
[tex]e) \ \bigg(\dfrac{4}{10}^{(2} \bigg)^{34} : \bigg(\dfrac{2}{5} \bigg)^{33} = \bigg(\dfrac{2}{5}\bigg)^{34} : \bigg(\dfrac{2}{5} \bigg)^{33} =\\[/tex]
[tex]= \bigg(\dfrac{2}{5} \bigg)^{34-33} = \bigg(\dfrac{2}{5} \bigg)^{1} = \bf \dfrac{2}{5}[/tex]
[tex]f) \ \bigg(\dfrac{4}{9}\bigg)^{34} : \bigg(\dfrac{2}{3} \bigg)^{65} = \bigg(\dfrac{2^2}{3^2}\bigg)^{34} : \bigg(\dfrac{2}{3} \bigg)^{65} = \bigg[\bigg(\dfrac{2}{3}\bigg)^{2}\bigg]^{35} : \bigg(\dfrac{2}{3} \bigg)^{65} = \\[/tex]
[tex]= \bigg(\dfrac{2}{3} \bigg)^{2\cdot35} : \bigg(\dfrac{2}{3} \bigg)^{65} = \bigg(\dfrac{2}{3} \bigg)^{70} : \bigg(\dfrac{2}{3} \bigg)^{65} = \bigg(\dfrac{2}{3} \bigg)^{70-65} \\[/tex]
[tex]= \bigg(\dfrac{2}{3} \bigg)^{5} = \dfrac{2^5}{3^5} = \bf\dfrac{32}{243}[/tex]
✍ Se utilizează regulile de calcul cu puteri:
[tex]\boxed{\boxed{\boldsymbol{a^{m} \cdot a^{n} = a^{m + n}}}}[/tex]
[tex]\boxed{\boxed{\boldsymbol{a^{m} : a^{n} = a^{m - n}} }}[/tex]
[tex]\boxed{\boxed{\boldsymbol{(a^{m})^{n} = a^{m \cdot n} }}}[/tex]
