Esercizio 10 – Espressione con i numeri razionali

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

    \[\dfrac{  \left(2-\dfrac{3}{5}-\dfrac{33}{15}\right)+\left(1+\dfrac{4}{3}-\dfrac{1}{2}:3-1+\dfrac{1}{6}\right)^3: \dfrac{5}{27}}{ \left(1-\dfrac{1}{4}\right):\dfrac{3}{8}+\dfrac{25}{32}: \left(1-\dfrac{1}{2}:4-\dfrac{3}{2}:6\right)^2}\]

 

Soluzione

Procediamo come segue

    \[\begin{aligned}		 	& 					\dfrac{  \left(2-\dfrac{3}{5}-\dfrac{33}{15}\right)+\left(1+\dfrac{4}{3}-\dfrac{1}{2}:3-1+\dfrac{1}{6}\right)^3: \dfrac{5}{27}}{ \left(1-\dfrac{1}{4}\right):\dfrac{3}{8}+\dfrac{25}{32}: \left(1-\dfrac{1}{2}:4-\dfrac{3}{2}:6\right)^2}= \\\\ 	& = 					\dfrac{  \dfrac{30-9-33}{15} + \left(\cancel{1}+\dfrac{4}{3}-\dfrac{1}{2}\cdot \dfrac{1}{3}-\cancel{1}+\dfrac{1}{6}\right)^3 \cdot \dfrac{27}{5}}{ \dfrac{4-1}{4}\cdot \dfrac{8}{3} + \dfrac{25}{32}: \left(1-\dfrac{1}{2}\cdot\dfrac{1}{4}-\dfrac{3}{2}\cdot\dfrac{1}{6}\right)^2} = \\\\ 	& = \dfrac{  -\dfrac{12}{15} + \left(\dfrac{4}{3}\cancel{-\dfrac{1}{6}}+\cancel{\dfrac{1}{6}}\right)^3 \cdot \dfrac{27}{5}}{ \dfrac{3}{4}\cdot \dfrac{8}{3} + \dfrac{25}{32}: \left(1-\dfrac{1}{8}-\dfrac{1}{4}\right)^2} = \\\\ 	& = \dfrac{  -\dfrac{\cancel{12}^{\;4}}{\cancel{15}_{\; 5}} + \left(\dfrac{4}{3}\right)^3 \cdot \dfrac{27}{5}}{ 2 + \dfrac{25}{32}: \left(\dfrac{8-1-2}{8}\right)^2} = \dfrac{  -\dfrac{4}{5} + \dfrac{64}{27}\cdot \dfrac{27}{5}}{ 2 + \dfrac{25}{32}: \left(\dfrac{5}{8}\right)^2} = \\\\ 	& = \dfrac{  -\dfrac{4}{5} + \dfrac{64}{\cancel{27}}\cdot \dfrac{\cancel{27}}{5}}{ 2 + \dfrac{25}{32}: \left(\dfrac{5}{8}\right)^2} = 	\dfrac{  -\dfrac{4}{5} + \dfrac{64}{5}}{ 2 + \dfrac{25}{32}: \dfrac{25}{64}} = \\\\ 	& = 	\dfrac{  \dfrac{64-4}{5} }{ 2 + \dfrac{25}{32}: \dfrac{25}{64}} = 	\dfrac{  \dfrac{60}{5} }{ 2 + \dfrac{\cancel{25}}{\cancel{32}_{\;1}}\cdot \dfrac{\cancel{64}^{\;2}}{\cancel{25}}} = \\\\ 	& = 	\dfrac{  12 }{ 2 + 2} = 3 \end{aligned}\]


Fonte: Algebra Blu con Statistica – Volume 1