[tex]$ \ $ Nu inteleg foarte bine ce cuprinde acea ultima putere, asa ca voi scrie \\ cele 2 variante posibile. \\ \\ \boxed{1} \quad \Big( \dfrac{x+1}{2x+1} \Big)^{-1^{-1}}=\Big( \dfrac{x+1}{2x+1} \Big)^{-1} = \dfrac{2x+1}{x+1} \\ \\\boxed{ -1^{-1} = \Big(-\dfrac{1}{1}\Big) ^{-1} = -\dfrac{1}{1} = -1} \\ \\ \\ \\ \boxed{2} \quad \Big[\Big( \dfrac{x+1}{2x+1} \Big)^{-1}\Big]^{-1} = \Big(\dfrac{x+1}{2x+1}\Big)^{-1\cdot(-1)} = \Big(\dfrac{x+1}{2x+1}\Big)^{1} = \\ \\ =\dfrac{x+1}{2x+1}[/tex]
