Imi poate rezolva si mie cineva acest ex va rog! Daca s-ar putea o rezolvare de clasa a 5-a

[tex]\it 3\dfrac{4}{15}:2\dfrac{1}{10}-\Big(\dfrac{2}{3}\Big)^3\cdot1\dfrac{31}{32}=\dfrac{49}{15}:\dfrac{21}{10}-\dfrac{8}{27}\cdot\dfrac{63}{32}=\\ \\ \\ =\dfrac{49}{15}\cdot\dfrac{10}{21}-\dfrac{8}{27}\cdot\dfrac{63}{32}=\dfrac{^{4)}14}{\ 9}\cdot\dfrac{^{3)}7}{\ 12}=\dfrac{56-21}{36}=\dfrac{35}{36}[/tex]