Frazioni algebriche – Esercizio 50
In questo cinquantesimo articolo sulle frazioni algebriche, presentiamo un espressione completamente risolta sull’argomento. Segnaliamo anche il precedente Frazioni algebriche – Esercizio 49 per ulteriore materiale sulle frazioni algebriche.
Buona lettura!
Esercizio 50 
. Semplificare la seguente espressione assumendo che siano verificate le condizioni di esistenza:
. Semplificare la seguente espressione assumendo che siano verificate le condizioni di esistenza:
![\[\dfrac{\dfrac{4}{a^{2n}-a^n-2} : \left(\dfrac{1}{a^{2n}-3a^n+2}+\dfrac{1}{a^{2n}-a^n-2} - \dfrac{2a^m}{a^m-a^{2n+m}}\right)}{\dfrac{a^n+1}{2a^n-2}: \left(\dfrac{2a^{2n}-3a^n+1}{a^{3n}-3a^{2n}+3a^n-1}-\dfrac{a^n}{a^{2n}-2a^n+1}-\dfrac{a^{n+m}}{a^m-a^{n+m}}\right)}.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-552e7220607ad7f5a8969f2211bb688d_l3.svg "Rendered by QuickLaTeX.com")
Svolgimento.
Possiamo scrivere
![\[\begin{aligned} & \dfrac{\dfrac{4}{a^{2n}-a^n-2} : \left(\dfrac{1}{a^{2n}-3a^n+2}+\dfrac{1}{a^{2n}-a^n-2} - \dfrac{2a^m}{a^m-a^{2n+m}}\right)}{\dfrac{a^n+1}{2a^n-2}: \left(\dfrac{2a^{2n}-3a^n+1}{a^{3n}-3a^{2n}+3a^n-1}-\dfrac{a^n}{a^{2n}-2a^n+1}-\dfrac{a^{n+m}}{a^m-a^{n+m}}\right)}=\\\\ & = \dfrac{\dfrac{4}{(a^n+1)(a^n-2)} : \left(\dfrac{1}{(a^n-1)(a^n-2)}+\dfrac{1}{(a^n+1)(a^n-2)} + \dfrac{2a^m}{a^m(a^{n}-1)(1+a^n)}\right)}{\dfrac{a^n+1}{2(a^n-1)}: \left(\dfrac{2a^{2n}-3a^n+1}{(a^{n}-1)^3}-\dfrac{a^n}{(a^{n}-1)^2}+\dfrac{a^{n+m}}{a^m(a^{n}-1)}\right)}=\\\\ \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-9e03b26c377bed2b8c8b44eb4f3f6480_l3.svg "Rendered by QuickLaTeX.com")
![\[\begin{aligned} & = \dfrac{\dfrac{4}{(a^n+1)(a^n-2)} : \left(\dfrac{1}{(a^n-1)(a^n-2)}+\dfrac{1}{(a^n+1)(a^n-2)} + \dfrac{2\cancel{a^m}}{\cancel{a^m}(a^{n}-1)(1+a^n)}\right)}{\dfrac{a^n+1}{2(a^n-1)}: \left(\dfrac{2a^{2n}-3a^n+1}{(a^{n}-1)^3}-\dfrac{a^n}{(a^{n}-1)^2}+\dfrac{a^{n}\cancel{a^m}}{\cancel{a^m}(a^{n}-1)}\right)}=\\\\ & = \dfrac{\dfrac{4}{(a^n+1)(a^n-2)} : \left(\dfrac{a^n+1+a^n-1+2a^n-4}{(a^n-1)(a^n-2)(a^n+1)}\right)}{\dfrac{a^n+1}{2(a^n-1)}: \left(\dfrac{2a^{2n}-3a^n+1-a^n(a^n-1)+a^n(a^n-1)^2}{(a^{n}-1)^3}\right)}=\\\\ & = \dfrac{\dfrac{4}{(a^n+1)(a^n-2)} : \left(\dfrac{4(a^n-1)}{(a^n-1)(a^n-2)(a^n+1)}\right)}{\dfrac{a^n+1}{2(a^n-1)}: \left(\dfrac{(a^{2n}-1)(a^n-1)}{(a^{n}-1)^3}\right)}=\\\\ & = \dfrac{\dfrac{4}{(a^n+1)(a^n-2)} \cdot \dfrac{(a^n-1)(a^n-2)(a^n+1)}{4(a^n-1)}}{\dfrac{a^n+1}{2(a^n-1)} \cdot \dfrac{(a^{n}-1)^3}{(a^{2n}-1)(a^n-1)}}=\\\\ & = \dfrac{1}{\dfrac{a^n+1}{2} \cdot \dfrac{(a^{n}-1)}{(a^{2n}-1)}}=\\\\ & = \dfrac{1}{\dfrac{a^n+1}{2} \cdot \dfrac{(a^{n}-1)}{(a^{n}-1)(a^n+1)}}=\\\\ & = 2. \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-a341d939f342a19a0ab0f5ba6c9d3bbe_l3.svg "Rendered by QuickLaTeX.com")