Найдите значение дроби предворительно сократив её 20/45+26/54, 14/24-39/90+15/100,...

$\frac{20}{45} + \frac{26}{54} = \frac{4}{9} + \frac{13}{27} = \frac{4*3}{9*3} + \frac{13}{27} = \frac{12+13}{27} = \frac{25}{27}$

$\frac{14}{24}- \frac{39}{90} + \frac{15}{100} = \frac{7}{12} - \frac{13}{30} + \frac{3}{20} =\frac{7*5}{12*5} - \frac{13*2}{30*2} + \frac{3*3}{20*3} = \frac{35-26+9}{60} = \frac{18}{60}$ $= \frac{3}{10}$

$\frac{45}{72} - \frac{33}{144} - \frac{20}{64} = \frac{5}{8} - \frac{11}{48}- \frac{5}{16} = \frac{5*6}{8*6} - \frac{11}{48}- \frac{5*3}{16*3} = \frac{30-11-15}{48} = \frac{4}{48} = \frac{1}{12}$