Esercizio. 
Semplificare la seguente espressione assumendo che siano verificate le condizioni di esistenza:
![\[\left\{ \left[\left(1-\dfrac{1}{x}\right):\left(1-\dfrac{1}{x^2}\right)\right]^2-\dfrac{x^2}{x^2+2x+1}\right\}:\dfrac{1}{x^2-1}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-a95a3ec6a0adfef6550c2adb2674ec74_l3.png "Rendered by QuickLaTeX.com")
Soluzione.
Sfruttando
Sfruttando
![\[A^2-B^2 = (A-B)(A+B)\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-00e6d43f9ef7f72fcc7c39293db5ed49_l3.png "Rendered by QuickLaTeX.com")
e
![\[A^2+2AB+B^2=(A+B)^2\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-57a310c072e86b584b02913abe7a12ff_l3.png "Rendered by QuickLaTeX.com")
possiamo scrivere
![\[\begin{aligned} & \left\{ \left[\left(1-\dfrac{1}{x}\right):\left(1-\dfrac{1}{x^2}\right)\right]^2-\dfrac{x^2}{x^2+2x+1}\right\}:\dfrac{1}{x^2-1} =\\\\ & = \left\{ \left[\left(\dfrac{x-1}{x}\right):\left(\dfrac{(x-1)(x+1)}{x^2}\right)\right]^2-\dfrac{x^2}{(x+1)^2}\right\}:\dfrac{1}{(x-1)(x+1)} =\\\\ & = \left\{ \dfrac{x^2}{(x+1)^2} -\dfrac{x^2}{(x+1)^2}\right\}:\dfrac{1}{(x-1)(x+1)} =\\\\ & = 0:\dfrac{1}{(x-1)(x+1)} =\\\\ & = 0 \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-cbda5b79ecc962993ea865e593d2668b_l3.png "Rendered by QuickLaTeX.com")
Fonte: Moduli di lineamenti di matematica N.Dodero – P.Baroncini – R.Manfredi