Esercizio. 
Semplificare la seguente espressione assumendo che siano verificate le condizioni di esistenza:
![\[\left[ \dfrac{1}{2a+a^{n+1}} : \dfrac{(-2a^n)^2}{4-a^{2n}} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{2a^n}\right)^3\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-8a513831ed3222e2570686bc3ef9d0c7_l3.png "Rendered by QuickLaTeX.com")
Soluzione.
Procediamo come segue
Procediamo come segue
![\[\begin{aligned} & \left[ \dfrac{1}{\underbrace{2a+a^{n+1}}_{\text{raccolgo $a$}}} : \dfrac{(-2a^n)^2}{\underbrace{4-a^{2n}}_{\text{differenza di quadrati}}} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{2a^n}\right)^3 = \\\\ & = \left[ \dfrac{1}{a(2+a^{n})} : \dfrac{4a^{2n}}{(2-a^n)(2+a^n)} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{8a^{3n}}\right) = \\\\ & = \left[ \dfrac{1}{a(2+a^{n})} \cdot \dfrac{(2-a^n)(2+a^n)}{4a^{2n}} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{8a^{3n}}\right) = \\\\ & = \left[ \dfrac{1}{a} \cdot \dfrac{2-a^n}{4a^{2n}} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{8a^{3n}}\right) = \\\\ & = \left[ \dfrac{2-a^n}{4a^{2n+1}} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{8a^{3n}}\right) = \\\\ & = - \dfrac{a^n}{4a^{2n+1}} : \left(-\dfrac{1}{8a^{3n}}\right) = \\\\ & = \dfrac{a^n}{4a^{2n+1}} \cdot 8a^{3n} = \\\\ & = \dfrac{8a^{4n}}{4a^{2n+1}} = \\\\ & = 2a^{4n-2n-1} =\\\\ & = 2a^{2n-1} \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-f59d7ea9fb2ca2764fe47b76952fb86e_l3.png "Rendered by QuickLaTeX.com")
Fonte: Moduli di lineamenti di matematica N.Dodero – P.Baroncini – R.Manfredi