[tex]\it \Big(-\dfrac{1}{2}\Big)^{-3} = \Big[\Big(-\dfrac{1}{2}\Big)^{-1}\Big]^3= \Big(-\dfrac{2}{1}\Big)^{3}=(-2)^3=-8\\ \\ \\ \Big(\dfrac{1}{3}\Big)^{-2}= \Big[\Big(\dfrac{1}{3}\Big)^{-1}\Big]^2= \Big(\dfrac{3}{1}\Big)^2= 3^2=9\\ \\ \\ 3^{-1}=\Big(\dfrac{3}{1}\Big)^{-1}=\dfrac{1}{3}\\ \\ \\ 0,3^{-1}=\Big (\dfrac{3}{10}\Big)^{-1}=\dfrac{10}{3}[/tex]
Acum, exercițiul devine:
[tex]\it -8\cdot\dfrac{1}{4}+9\cdot\dfrac{1}{3}+\dfrac{10}{3}\cdot\dfrac{1}{5}=-2+3+\dfrac{2}{3}=1+\dfrac{2}{3}=\dfrac{5}{3}[/tex]
