Răspuns:
Respectăm regulile de calcul cu puteri:
[tex]c) = (3 \cdot 16 - 81)^2 \cdot \dfrac{2}{63} + \dfrac{125-25}{16-4} = (48 - 81)^2 \cdot \dfrac{2}{63} + \dfrac{100}{12}^{(4} \\[/tex]
[tex]= (-33)^2 \cdot \dfrac{2}{63} + \dfrac{25}{3} = \dfrac{1089 \cdot 2}{63}^{(3} + \dfrac{^{7)} 25}{3} = \dfrac{726}{21} + \dfrac{175}{21} \\[/tex]
[tex]= \dfrac{726+175}{21} = \bf\dfrac{901}{21}[/tex]
[tex]d) = \bigg(-\dfrac{2+1}{2}\bigg)^{26} \cdot \bigg(\dfrac{15}{10}^{(5} \bigg)^{2 \cdot 3} \cdot \bigg(\dfrac{4}{9} \bigg)^{16} = \bigg(-\dfrac{3}{2}\bigg)^{26} \cdot \bigg(\dfrac{3}{2}\bigg)^{6} \cdot \bigg(\dfrac{2^2}{3^2}\bigg)^{16} \\[/tex]
[tex]= \bigg(\dfrac{3}{2}\bigg)^{26} \cdot \bigg(\dfrac{3}{2}\bigg)^{6} \cdot \bigg[\bigg(\dfrac{2}{3}\bigg)^2\bigg]^{16} = \bigg(\dfrac{3}{2}\bigg)^{26 + 6} \cdot \bigg(\dfrac{2}{3}\bigg)^{2 \cdot 16} = \bigg(\dfrac{3}{2}\bigg)^{32} \cdot \bigg(\dfrac{2}{3}\bigg)^{32}\\[/tex]
[tex]= \bigg(\dfrac{3}{2}\bigg)^{32} \cdot \bigg(\dfrac{3}{2}\bigg)^{-32} = \bigg(\dfrac{3}{2}\bigg)^{32-32} = \bigg(\dfrac{3}{2}\bigg)^{0} = \bf 1\\[/tex]
✍ Reguli de calcul cu puteri:
[tex]\boxed{ \boldsymbol{ a^{m} \cdot a^{n} = a^{m + n} }}[/tex]
[tex]\boxed{ \boldsymbol{ a^{m} : a^{n} = a^{m - n} }}[/tex]
[tex]\boxed{ \boldsymbol{ (a^{m})^{n} = a^{m \cdot n} }}[/tex]
[tex]\boxed{ \boldsymbol{ a^{m} \cdot b^{m} = (a \cdot b)^{m} }}[/tex]
[tex]\boxed{\boldsymbol{a^{-m}=\dfrac{1}{a^{m} }}}[/tex]
