Esercizio 3 – Espressione con i numeri razionali

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

    \[\left\{ \left[ \left(3-\dfrac{13}{11} \right) : \left(\dfrac{3}{2}-\dfrac{4}{11} \right)\right] - \dfrac{3}{5}\right\} \cdot \dfrac{8}{3}-\left(\dfrac{4}{3}: \dfrac{1}{3}-4+\dfrac{16}{5}\right)\]

 

Soluzione

Procediamo come segue

    \[\begin{aligned} & 			\left\{ \left[ \left(3-\dfrac{13}{11} \right) : \left(\dfrac{3}{2}-\dfrac{4}{11} \right)\right] - \dfrac{3}{5}\right\} \cdot \dfrac{8}{3}-\left(\dfrac{4}{3}: \dfrac{1}{3}-4+\dfrac{16}{5}\right) = \\\\ & = 			\left\{ \left[ \left(\dfrac{33-13}{11} \right) : \left(\dfrac{33-8}{22}\right)\right] - \dfrac{3}{5}\right\} \cdot \dfrac{8}{3}-\left(\dfrac{4}{3}\cdot 3-4+\dfrac{16}{5}\right) = \\\\ & = 			\left\{ \left[ \dfrac{20}{11}  : \dfrac{25}{22}\right] - \dfrac{3}{5}\right\} \cdot \dfrac{8}{3}-\left(\dfrac{4}{\cancel{3}}\cdot \cancel{3}-4+\dfrac{16}{5}\right) = \\\\  & = 			\left\{ \left[ \dfrac{\cancel{20}^{\; 4}}{\cancel{11}_{\; 1}}  \cdot \dfrac{\cancel{22}^{\; 2}}{\cancel{25}_{\; 5}}\right] - \dfrac{3}{5}\right\} \cdot \dfrac{8}{3}-\dfrac{16}{5} = \\\\  & = 			\left\{ \dfrac{8}{5} - \dfrac{3}{5}\right\} \cdot \dfrac{8}{3}-\dfrac{16}{5} = \\\\  & = 	1 \cdot \dfrac{8}{3}-\dfrac{16}{5} = \\\\  & =\dfrac{8}{3}-\dfrac{16}{5} = \dfrac{40-48}{15} = - \dfrac{8}{15} \end{aligned}\]


Fonte: Algebra Blu con Statistica – Volume 1
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