Esercizio 5 – Espressione con i numeri razionali

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

    \[\dfrac{-\left(-\dfrac{1}{3}+\dfrac{2}{5}\right)}{-\dfrac{3}{5}-\left(-\dfrac{2}{3}\right)} + \dfrac{\dfrac{5}{6}+\left(-\dfrac{1}{3}\right)}{-\dfrac{3}{4}+\dfrac{5}{12}}\]

 

Soluzione

Procediamo come segue

    \[\begin{aligned} \dfrac{-\left(-\dfrac{1}{3}+\dfrac{2}{5}\right)}{-\dfrac{3}{5}-\left(-\dfrac{2}{3}\right)} + \dfrac{\dfrac{5}{6}+\left(-\dfrac{1}{3}\right)}{-\dfrac{3}{4}+\dfrac{5}{12}} & = \dfrac{-\left(\dfrac{-5+6}{15}\right)}{-\dfrac{3}{5}+\dfrac{2}{3}} + \dfrac{\dfrac{5}{6}-\dfrac{1}{3}}{\dfrac{-9+5}{12}} = \\\\ & = \dfrac-{\dfrac{1}{15}}{\dfrac{-9+10}{15}} + \dfrac{\dfrac{5-2}{6}}{-\dfrac{4}{12}} =		\dfrac{-\dfrac{1}{15}}{\dfrac{1}{15}} + \dfrac{\dfrac{\cancel{3}^{\;1}}{\cancel{6}_{\;2}}}{-\dfrac{\cancel{4}^{\;1}}{\cancel{12}_{\; 3}}} =\\\\ & =	-\dfrac{1}{\cancel{15}_{\;1}} \cdot \cancel{15}^{\;1} + \dfrac{1}{2} \cdot (-3) =- 1 -\dfrac{3}{2} = \dfrac{-2-3}{2} = -\dfrac{5}{2} 	\end{aligned}\]


Fonte: Algebra Blu con Statistica – Volume 1