Esercizio 2 – Espressione con i numeri razionali

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

    \[\left\{ \left[ \left(\dfrac{1}{3}-\dfrac{4}{7}\right): \left(\dfrac{2}{7}-\dfrac{1}{2}\right) \cdot \dfrac{3}{5}-\dfrac{2}{3} \right] : 2 \right\} \cdot \dfrac{1}{4}-\dfrac{2}{3}+1\]

 

Soluzione

Procediamo come segue

    \[\begin{aligned} & \left\{ \left[ \left(\dfrac{1}{3}-\dfrac{4}{7}\right): \left(\dfrac{2}{7}-\dfrac{1}{2}\right) \cdot \dfrac{3}{5}-\dfrac{2}{3} \right] : 2 \right\} \cdot \dfrac{1}{4}-\dfrac{2}{3}+1 = \\\\ & = \left\{ \left[ \left(\dfrac{7-12}{21}\right): \left(\dfrac{4-7}{14}\right) \cdot \dfrac{3}{5}-\dfrac{2}{3} \right] \cdot \dfrac{1}{2} \right\} \cdot \dfrac{1}{4}-\dfrac{2}{3}+1 = \\\\  & = \left\{ \left[ \left(-\dfrac{5}{21}\right): \left(-\dfrac{3}{14}\right) \cdot \dfrac{3}{5}-\dfrac{2}{3} \right] \cdot \dfrac{1}{2} \right\} \cdot \dfrac{1}{4}-\dfrac{2}{3}+1 = \\\\  & = \left\{ \left[ \left(-\dfrac{5}{21}\right) \cdot \left(-\dfrac{14}{3}\right) \cdot \dfrac{3}{5}-\dfrac{2}{3} \right] \cdot \dfrac{1}{2} \right\} \cdot \dfrac{1}{4}-\dfrac{2}{3}+1 = \\\\  & = \left\{ \left[ \left(-\dfrac{\cancel{5}}{\cancel{21}_{\; 3}}\right) \cdot \left(-\dfrac{\cancel{14}^{\; 2}}{\cancel{3}}\right) \cdot \dfrac{\cancel{3}}{\cancel{5}}-\dfrac{2}{3} \right] \cdot \dfrac{1}{2} \right\} \cdot \dfrac{1}{4}-\dfrac{2}{3}+1 = \\\\  & = \left\{ \left[ \dfrac{2}{3} -\dfrac{2}{3} \right] \cdot \dfrac{1}{2} \right\} \cdot \dfrac{1}{4}-\dfrac{2}{3}+1 = \\\\ & = \left\{ 0 \cdot \dfrac{1}{2} \right\} \cdot \dfrac{1}{4}-\dfrac{2}{3}+1 = -\dfrac{2}{3}+1 = \dfrac{1}{3} \end{aligned}\]


Fonte: Algebra Blu con Statistica – Volume 1