Espressioni con i polinomi – Esercizio 4

Polinomi: Espressioni e Problemi

Home » Espressioni con i polinomi – Esercizio 4
Generic selectors
Exact matches only
Search in title
Search in content
Post Type Selectors
post
page

Esercizio.  (\bigstar\largewhitestar\largewhitestar) Semplificare la seguente espressione

    \[(x+1)\left(-0.5 x+\dfrac{1}{2}\right)-\left(\dfrac{1}{2}x+1\right)\left(-\dfrac{1}{3}x-0.\overline{3}\right)\]

 

Soluzione

Facciamo i calcoli:

    \[\begin{aligned} &	(x+1)\left(-0.5 x+\dfrac{1}{2}\right)-\left(\dfrac{1}{2}x+1\right)\left(-\dfrac{1}{3}x-0.\overline{3}\right) = \\\\ & = (x+1)\left(-\dfrac{1}{2} x+\dfrac{1}{2}\right)-\left(\dfrac{1}{2}x+1\right)\left(-\dfrac{1}{3}x-\dfrac{1}{3}\right) = \\\\ & = -\dfrac{1}{2}x^2 -\dfrac{1}{2}x + \dfrac{1}{2}x + \dfrac{1}{2} - \left( - \dfrac{1}{6}x^2 - \dfrac{1}{3}x - \dfrac{1}{6}x^2 - \dfrac{1}{3}x\right) = \\\\ & = -\dfrac{1}{2}x^2+\dfrac{1}{2} - \left( - \dfrac{2}{6} x^2 - \dfrac{2}{3}x\right) = \\\\ & = -\dfrac{1}{2}x^2+\dfrac{1}{2} - \left( - \dfrac{1}{3} x^2 - \dfrac{2}{3}x\right) = \\\\ & = -\dfrac{1}{2}x^2+\dfrac{1}{2} +\dfrac{1}{3} x^2 + \dfrac{2}{3}x = \\\\ & = - \dfrac{1}{6}x^2 + \dfrac{2}{3}x + \dfrac{1}{2} \end{aligned}\]


Fonte: Algebra.blu 2 – Zanichelli