Esercizio 9 – Espressione con i numeri razionali

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

    \[\left\{ \left[ \left(\dfrac{4}{3}-\dfrac{1}{4} \right) \cdot \left(-\dfrac{2}{13}+3\right) -\dfrac{30}{12}\right] : \dfrac{7}{4}-\dfrac{2}{3}\right\} \cdot \dfrac{3}{5}-\dfrac{2}{5}+\dfrac{9^3}{36}: \dfrac{3^4}{4}\]

 

Soluzione

Procediamo come segue

    \[\begin{aligned}		 	& 			\left\{ \left[ \left(\dfrac{4}{3}-\dfrac{1}{4} \right) \cdot \left(-\dfrac{2}{13}+3\right) -\dfrac{30}{12}\right] : \dfrac{7}{4}-\dfrac{2}{3}\right\} \cdot \dfrac{3}{5}-\dfrac{2}{5}+\dfrac{9^3}{4\cdot 9}: \dfrac{9^2}{4} = \\\\ 	& = 			\left\{ \left[ \dfrac{16-3}{12} \cdot \dfrac{-2+39}{13} -\dfrac{\cancel{30}^{\; 5}}{\cancel{12}_{\;2}}\right] \cdot \dfrac{4}{7}-\dfrac{2}{3}\right\} \cdot \dfrac{3}{5}-\dfrac{2}{5}+\dfrac{\cancel{9^3}}{\cancel{4}^{\;1}} \cdot \dfrac{\cancel{4}^{\;1}}{\cancel{9^3}} = \\\\ 	& = \left\{ \left[ \dfrac{\cancel{13}}{12} \cdot \dfrac{37}{\cancel{13}} -\dfrac{5}{2}\right] \cdot \dfrac{4}{7}-\dfrac{2}{3}\right\} \cdot \dfrac{3}{5}-\dfrac{2}{5}+1 = \\\\ 	& = \left\{ \left[ \dfrac{37}{12} -\dfrac{5}{2}\right] \cdot \dfrac{4}{7}-\dfrac{2}{3}\right\} \cdot \dfrac{3}{5}-\dfrac{2}{5}+1 = \\\\ 	& = \left\{ \dfrac{37-30}{12} \cdot \dfrac{4}{7}-\dfrac{2}{3}\right\} \cdot \dfrac{3}{5}-\dfrac{2}{5}+1 = \\\\ 	& = \left\{ \dfrac{\cancel{7}}{\cancel{12}_{\;3}} \cdot \dfrac{\cancel{4}^{\;1}}{\cancel{7}}-\dfrac{2}{3}\right\} \cdot \dfrac{3}{5}-\dfrac{2}{5}+1 = \\\\ 	& = \left\{ \dfrac{\cancel{7}}{\cancel{12}_{\;3}} \cdot \dfrac{\cancel{4}^{\;1}}{\cancel{7}}-\dfrac{2}{3}\right\} \cdot \dfrac{3}{5}-\dfrac{2}{5}+1 = \\\\ 	& = \left\{ \dfrac{1}{3} - \dfrac{2}{3}\right\} \cdot \dfrac{3}{5}-\dfrac{2}{5}+1 = \\\\ 	& = - \dfrac{1}{\cancel{3}} \cdot \dfrac{\cancel{3}}{5}-\dfrac{2}{5}+1 = \\\\ 	& = - \dfrac{1}{5} -\dfrac{2}{5}+1 = \dfrac{2}{5}	 \end{aligned}\]


Fonte: Algebra Blu con Statistica – Volume 1