Espressione che è stata assegnata su classroom:
Esercizio 1 Calcola il valore della seguente espressione applicando le regole finora studiate:
$\dfrac{\left(\dfrac{11}{3}+\dfrac{7}{9}\right):[(-2^{-3}]^{-1}+\dfrac{1}{5}}{\left[\left(\dfrac{11}{2}-\dfrac{3}{5}\right)-\dfrac{7}{5}\right]:\left[\left(1+\dfrac{1}{3}\right)\cdot\left(\dfrac{7}{8}\right)\right]-3}\cdot\dfrac{\left(1+\dfrac{1}{2}\right)^2}{\left(\dfrac{6}{7}-\dfrac{1}{2}\right)^{-2}}=$
$=\dfrac{\left(\dfrac{33+7}{9}\right):[-2]^{3}+\dfrac{1}{5}}{\left[\dfrac{55-6}{10}-\dfrac{7}{5}\right]:\left[\left(\dfrac{3+1}{3}\right)\cdot\left(\dfrac{7}{8}\right)\right]-3}\cdot\dfrac{\left(\dfrac{3}{2}\right)^2}{\left(\dfrac{12-7}{14}\right)^{-2}}=$
$=\dfrac{\left(\dfrac{40}{9}\right)\cdot\left(-\dfrac{1}{8}\right)+\dfrac{1}{5}}{\left[\dfrac{49}{10}-\dfrac{7}{5}\right]:\left[\left(\dfrac{4}{3}\right)\cdot\left(\dfrac{7}{8}\right)\right]-3}\cdot\dfrac{\dfrac{9}{4}}{\left(\dfrac{5}{14}\right)^{-2}}=\dfrac{-\dfrac{5}{9}+\dfrac{1}{5}}{\left[\dfrac{49-14}{10}\right]:\left[\dfrac{7}{6}\right]-3}\cdot\dfrac{\dfrac{9}{4}}{\left(\dfrac{14}{5}\right)^{2}}=$
$=\dfrac{-\dfrac{5}{9}+\dfrac{1}{5}}{\left[\dfrac{35}{10}\right]\cdot\left[\dfrac{6}{7}\right]-3}\cdot\dfrac{\dfrac{9}{4}}{\left(\dfrac{14}{5}\right)^{2}}=\dfrac{-\dfrac{5}{9}+\dfrac{1}{5}}{\left[\dfrac{7}{2}\right]\cdot\left[\dfrac{6}{7}\right]-3}\cdot\dfrac{\dfrac{9}{4}}{\left(\dfrac{14}{5}\right)^{2}}=\dfrac{-\dfrac{5}{9}+\dfrac{1}{5}}{3-3}\cdot\dfrac{\dfrac{9}{4}}{\left(\dfrac{14}{5}\right)^{2}}=$
$=\dfrac{-\dfrac{5}{9}+\dfrac{1}{5}}{0}\cdot\dfrac{\dfrac{9}{4}}{\left(\dfrac{14}{5}\right)^{2}}=$ priva di significato!
Dall'ultima espressione si evince che è impossibile perché si materializza $0$ a denominatore di una frazione e di conseguenza tutta l'espressione è priva di significato.