Frazioni algebriche – Esercizio 15

Frazioni algebriche

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Esercizio.  (\bigstar\bigstar\largewhitestar)

Semplificare la seguente espressione assumendo che siano verificate le condizioni di esistenza:

    \[\left[ \dfrac{1}{2a+a^{n+1}} : \dfrac{(-2a^n)^2}{4-a^{2n}} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{2a^n}\right)^3\]

 

Soluzione. 
Procediamo come segue

    \[\begin{aligned} & \left[ \dfrac{1}{\underbrace{2a+a^{n+1}}_{\text{raccolgo $a$}}} : \dfrac{(-2a^n)^2}{\underbrace{4-a^{2n}}_{\text{differenza di quadrati}}} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{2a^n}\right)^3 = \\\\ & = \left[ \dfrac{1}{a(2+a^{n})} : \dfrac{4a^{2n}}{(2-a^n)(2+a^n)} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{8a^{3n}}\right) = \\\\ & = \left[ \dfrac{1}{a(2+a^{n})} \cdot \dfrac{(2-a^n)(2+a^n)}{4a^{2n}} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{8a^{3n}}\right) = \\\\ & = \left[ \dfrac{1}{a} \cdot \dfrac{2-a^n}{4a^{2n}} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{8a^{3n}}\right) = \\\\ & = \left[ \dfrac{2-a^n}{4a^{2n+1}} - \dfrac{1}{2a^{2n+1}} \right]: \left(-\dfrac{1}{8a^{3n}}\right) = \\\\ & = - \dfrac{a^n}{4a^{2n+1}} : \left(-\dfrac{1}{8a^{3n}}\right) = \\\\ & = \dfrac{a^n}{4a^{2n+1}} \cdot 8a^{3n} = \\\\ & = \dfrac{8a^{4n}}{4a^{2n+1}} = \\\\ & = 2a^{4n-2n-1} =\\\\ & = 2a^{2n-1} \end{aligned}\]

 


Fonte: Moduli di lineamenti di matematica N.Dodero – P.Baroncini – R.Manfredi