Esercizio 1 – Espressione con i numeri razionali

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

    \[\dfrac{1}{2} + \dfrac{4}{3}-\left[\dfrac{1}{2}+\dfrac{2}{3}+\left(1+\dfrac{1}{4}\right) \cdot \dfrac{12}{5} \right] \cdot \dfrac{1}{3} + \left(\dfrac{1}{9}: \dfrac{3}{4}\right)-\dfrac{1}{27}\]

 

Soluzione

Procediamo come segue

    \[\begin{aligned} 	& \dfrac{1}{2} + \dfrac{4}{3}-\left[\dfrac{1}{2}+\dfrac{2}{3}+\left(1+\dfrac{1}{4}\right) \cdot \dfrac{12}{5} \right] \cdot \dfrac{1}{3} + \left(\dfrac{1}{9}: \dfrac{3}{4}\right)-\dfrac{1}{27} = \\\\ 	& = \dfrac{3+8}{6} -\left[\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{\cancel{5}}{\cancel{4}} \cdot \dfrac{\cancel{12}^{\;3}}{\cancel{5}} \right] \cdot \dfrac{1}{3} + \dfrac{4-1}{27} = \\\\ 	& = \dfrac{11}{6}-\left[\dfrac{1}{2}+\dfrac{2}{3} + 3\right] \cdot \dfrac{1}{3} + \dfrac{\cancel{3}}{\cancel{27}_{\; 9}}= \\\\ 	& = \dfrac{11}{6}-\dfrac{3+4+18}{6}\cdot \dfrac{1}{3} + \dfrac{1}{9} = \\ \\ 	& = \dfrac{11}{6}-\dfrac{25}{6}\cdot \dfrac{1}{3} + \dfrac{1}{9} = \dfrac{11}{6}-\dfrac{25}{18} + \dfrac{1}{9} = \\\\ 	& = \dfrac{33-25+2}{18} = \dfrac{10}{18} = \dfrac{5}{9} \end{aligned}\]


Fonte: Algebra Blu con Statistica – Volume 1