1) 29 49/53-8 49/53=2) (20 16/25+13 9/25)- (23 4/14+7 13/14)

Решите задачу:

$\displaystyle \boxed{1.} 29 \frac{49}{53}-8 \frac{49}{53}=(29-8)+( \frac{49}{53}- \frac{49}{53})=29-8+0=21$

$\displaystyle \boxed{2.} (20 \frac{16}{25}+13 \frac{9}{25})-(23 \frac{4}{14}+ 7 \frac{3}{14})=33 \frac{25}{25} - 30 \frac{17}{14}=$
$\displaystyle = 34-31 \frac{3}{14}= 33 \frac{14}{14}-31 \frac{3}{14}=2 \frac{11}{14}$