[math](\frac{7}{3}+\frac{1}{14}) \cdot (\frac{9}{2} - \frac{3}{5}) : (\frac{1}{6} - \frac{5}{4}) + [(\frac{3}{10} - \frac{2}{5} + 1) - (-\frac{2}{15} + \frac{1}{2})] \cdot \frac{2}{3} + (\frac{1}{4} + \frac{1}{3})[/math]
Soluzione
[math]\frac{98+3}{42} \cdot \frac{45-6}{10} : \frac{2-15}{12} + [\frac{3-4+10}{10}-\frac{-4+15}{30}] \cdot \frac{2}{3} + \frac{3+4}{12}=[/math]
[math]\frac{101}{42} \cdot \frac{39}{10} : \frac{-13}{12} + [\frac{9}{10} - \frac{11}{30}] \cdot \frac{2}{3} + \frac{7}{12}=[/math]
[math]-\frac{101}{7} \cdot \frac{3}{5} \cdot \frac{1}{1} + \frac{27-11}{30} \cdot \frac{2}{3} + \frac{7}{12}=[/math]
[math]-\frac{303}{35} + \frac{16}{30} \cdot \frac{2}{3} + \frac{7}{12}=[/math]
[math]-\frac{303}{35} + \frac{16}{15} \cdot \frac{1}{3} + \frac{7}{12}=[/math]
[math]-\frac{303}{35} + \frac{16}{45} + \frac{7}{12}=[/math]
[math]\frac{-10908+448+735}{1260} = \frac{-9725}{1260} = \frac{-1945}{252}[/math]