Răspunsuri

2014-03-26T21:54:38+02:00
\left(\dfrac32\right)^{-1}=\dfrac{1}{\dfrac32}=1\cdot\dfrac23=\dfrac23
Pentru al doilea am mai dat o rezolvare, asa ca acum dau alta:

\left(\dfrac14\right)^{-2}=\dfrac{1^{-2}}{4^{-2}}=\dfrac{1}{\dfrac{1}{4^2}}=1\cdot\dfrac{4^2}{1}=16