Frazioni algebriche – Esercizio 14

Frazioni algebriche

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Esercizio.  (\bigstar\largewhitestar\largewhitestar)

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

    \[\left(\dfrac{x^2}{1-3x+3x^2-x^3}-\dfrac{x}{x^2-2x+1}+\dfrac{2}{x-1}\right) \cdot \dfrac{x^2-2x+1}{(3x-2)(x+1)}\]

 

Soluzione. 
Procediamo come segue

    \[\begin{aligned} & \left(\dfrac{x^2}{\underbrace{1-3x+3x^2-x^3}_{\text{cubo di binomio}}}-\dfrac{x}{\underbrace{x^2-2x+1}_{\text{quadrato di binomio}}}+\dfrac{2}{x-1}\right) \cdot \dfrac{\overbrace{x^2-2x+1}^{\text{quadrato di binomio}}}{(3x-2)(x+1)} = \\ & = \left(\dfrac{x^2}{(1-x)^3}-\dfrac{x}{(x-1)^2}+\dfrac{2}{x-1}\right) \cdot \dfrac{(x-1)^2}{(3x-2)(x+1)} = \\ & = \left(-\dfrac{x^2}{(x-1)^3}-\dfrac{x}{(x-1)^2}+\dfrac{2}{x-1}\right) \cdot \dfrac{(x-1)^2}{(3x-2)(x+1)} = \\ & = \dfrac{-x^2-x(x-1)+2(x-1)^2}{(x-1)^3} \cdot \dfrac{(x-1)^2}{(3x-2)(x+1)} = \\\\ & = \dfrac{-3x+2}{(x-1)^3} \cdot \dfrac{(x-1)^2}{(3x-2)(x+1)} = \\\\ & = -\dfrac{\cancel{3x-2}}{(x-1)^{\cancel{3}}} \cdot \dfrac{\cancel{(x-1)^2}}{\cancel{(3x-2)}(x+1)} =\\\\ & = - \dfrac{1}{(x-1)(x+1)} = \\\\ & = - \dfrac{1}{x^2-1} = \\\\ & = \dfrac{1}{1-x^2} \end{aligned}\]

 


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