Упр.1066 ГДЗ Зубарева Мордкович 6 класс ФГОС (Математика)

а) (27 3/8-21 7/20)/((3 4/7-1 23/28)-(1 47/65-29/130) );
б) (4 2/5-3 2/4+8 7/15-8 7/60)/(2 3/4-4 1/4).

а)

$$ \frac{27\frac{3}{8}-21\frac{7}{20}}{\left(3\frac{4}{7}-1\frac{23}{28}\right)-\left(1\frac{47}{65}-\frac{29}{130}\right)} = \frac{27\frac{15}{40}-21\frac{14}{40}}{\left(3\frac{16}{28}-1\frac{23}{28}\right)-\left(1\frac{94}{130}-\frac{29}{130}\right)} $$
$$ = \frac{6\frac{1}{40}}{\left(2\frac{44}{28}-1\frac{23}{28}\right)-1\frac{65}{130}} = \frac{\frac{241}{40}}{1\frac{21}{28}-1\frac{1}{2}} = \frac{\frac{241}{40}}{1\frac{3}{4}-1\frac{1}{2}} = \frac{\frac{241}{40}}{\frac{1}{4}} $$
$$ =\frac{241}{40}\cdot 4=\frac{241}{10}=24{,}1. $$

б)

$$ \frac{4\frac{2}{5}-3\frac{2}{4}+8\frac{7}{15}-8\frac{7}{60}}{2\frac{3}{4}-4\frac{1}{4}} = \frac{4\frac{24}{60}-3\frac{30}{60}+8\frac{28}{60}-8\frac{7}{60}}{2\frac{3}{4}-3\frac{5}{4}} $$
$$ = \frac{3\frac{84}{60}-3\frac{30}{60}+\frac{21}{60}}{-1\frac{1}{2}} = \frac{\frac{54}{60}+\frac{21}{60}}{-\frac{3}{2}} = \frac{\frac{75}{60}}{-\frac{3}{2}} = \frac{5}{4}\cdot\left(-\frac{2}{3}\right) = -\frac{5}{6}. $$

Ответ

а) $$24{,}1$$; б) $$-\frac{5}{6}$$.