Упр.635 Часть 2 ГДЗ Дорофеев Петерсон 5 класс (Математика)

Задача

1) (12 4/5 · 3 3/4 — 4 4/11 · 4 1/8)/(11 2/3 · 2 4/7 : 4 1/4);
(12 4/5•3 3/4-4 4/11•4 1/8)/(11 2/3•2 4/7 :4 1/4)=((12•5+4)/5•(3•4+3)/4-(4•11+4)/11•(4•8+1)/8)/((11•3+2)/3•(2•7+4)/7 :(4•4+1)/4)=(64/5•15/4-48/11•33/8)/(35/3•18/7 :17/4)=((64•15)/(5•4)-(48•33)/(11•8))/((35•18)/(3•7)•4/17)=
=((4•16•3•5)/(5•4)-(6•8•3•11)/(11•8))/((5•7•3•6•4)/(3•7•17))=(48-18)/(120/17)=30/(120/17)=30:120/17=30•17/120=(30•17)/120=(30•17)/(30•4)=
=17/4=4 1/4
2) (9 4/5 : 4 2/3 + 3 1/3 : 1 1/3)/(5 3/4 : 2 1/2);
(9 4/5 :4 2/3+3 1/3 :1 1/3)/(5 3/4 :2 1/2)=((9•5+4)/5 :(4•3+2)/3+(3•3+1)/3 :(1•3+1)/3)/((5•4+3)/4 :(2•2+1)/2)=(49/5 :14/3+10/3 :4/3)/(23/4 :5/2)=(49/5•3/14+10/3•3/4)/(23/4•2/5)=
=((49•3)/(5•14)+(10•3)/(3•4))/((23•2)/(2•2•5))=((7•7•3)/(5•2•7)+(2•5•3)/(3•2•2))/(23/10)=(21/10+5/2)/(23/10)=(21/10+(5•5)/(2•5))/(23/10)=(21/10+25/10)/(23/10)=((21+25)/10)/(23/10)=(46/10)/(23/10)=
=46/10 :23/10=46/10•10/23=(46•10)/(10•23)=(2•23•10)/(10•23)=2/1=2
3) ((1 1/2 + 2 2/3 + 3 3/4) · 3 3/5 : 2 1/2)/(14 — 15 1/8 : 2 1/5).
((1 1/2+2 2/3+3 3/4)•3 3/5 :2 1/2)/(14-15 1/8 :2 1/5)=((1 (1•6)/(2•6)+2 (2•4)/(3•4)+3 (3•3)/(4•3))•(3•5+3)/5 :(2•2+1)/2)/(14-(15•8+1)/8 :(2•5+1)/5)=((1 6/12+2 8/12+3 9/12)•18/5 :5/2)/(14-121/8 :11/5)=
=((1+2+3+(6+8+9)/12)•18/5•2/5)/(14-121/8•5/11)=((6+23/12)•(18•2)/(5•5))/(14-(121•5)/(8•11))=((6+1 11/12)•(18•2)/(5•5))/(14-(11•11•5)/(8•11))=(7 11/12•(18•2)/(5•5))/(14-55/8)=((7•12+11)/12•(18•2)/(5•5))/(14-6 7/8)=
=(95/12•(18•2)/(5•5))/(13 8/8-6 7/8)=((95•18•2)/(12•5•5))/(13-6+(8-7)/8)=((5•19•3•6•2)/(2•6•5•5))/(7+1/8)=(57/5)/(7 1/8)=(57/5)/((7•8+1)/8)=(57/5)/(57/8)=57/5 :57/8=
=57/5•8/57=(57•8)/(5•57)=8/5=1 3/5

Подробный ответ

1) Преобразуем смешанные числа в неправильные дроби и выполняем действия по порядку:

$$ \frac{12\frac{4}{5}\cdot 3\frac{3}{4}-4\frac{4}{11}\cdot 4\frac{1}{8}}{11\frac{2}{3}\cdot 2\frac{4}{7}:4\frac{1}{4}} = \frac{\frac{64}{5}\cdot \frac{15}{4}-\frac{48}{11}\cdot \frac{33}{8}}{\frac{35}{3}\cdot \frac{18}{7}:\frac{17}{4}} $$

$$ \frac{\frac{64\cdot 15}{5\cdot 4}-\frac{48\cdot 33}{11\cdot 8}}{\frac{35\cdot 18}{3\cdot 7}\cdot \frac{4}{17}} = \frac{48-18}{\frac{120}{17}} = 30:\frac{120}{17} = 30\cdot \frac{17}{120} = \frac{17}{4} = 4\frac{1}{4} $$

2) Аналогично:

$$ \frac{9\frac{4}{5}:4\frac{2}{3}+3\frac{1}{3}:1\frac{1}{3}}{5\frac{3}{4}:2\frac{1}{2}} = \frac{\frac{49}{5}:\frac{14}{3}+\frac{10}{3}:\frac{4}{3}}{\frac{23}{4}:\frac{5}{2}} $$

$$ \frac{\frac{49}{5}\cdot \frac{3}{14}+\frac{10}{3}\cdot \frac{3}{4}}{\frac{23}{4}\cdot \frac{2}{5}} = \frac{\frac{21}{10}+\frac{5}{2}}{\frac{23}{10}} = \frac{\frac{21}{10}+\frac{25}{10}}{\frac{23}{10}} = \frac{\frac{46}{10}}{\frac{23}{10}} = \frac{46}{10}:\frac{23}{10} = 2 $$

3) Сначала вычислим сумму в числителе и произведение в знаменателе:

$$ \frac{\left(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}\right)\cdot 3\frac{3}{5}:2\frac{1}{2}}{14-15\frac{1}{8}:2\frac{1}{5}} $$

$$ 1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4} = \frac{3}{2}+\frac{8}{3}+\frac{15}{4} = \frac{18+32+45}{12} = \frac{95}{12} $$

$$ 3\frac{3}{5}:2\frac{1}{2} = \frac{18}{5}:\frac{5}{2} = \frac{18}{5}\cdot \frac{2}{5} = \frac{36}{25} $$

$$ 15\frac{1}{8}:2\frac{1}{5} = \frac{121}{8}:\frac{11}{5} = \frac{121}{8}\cdot \frac{5}{11} = \frac{55}{8} $$

Тогда
$$ \frac{\frac{95}{12}\cdot \frac{36}{25}}{14-\frac{55}{8}} = \frac{\frac{57}{5}}{7\frac{1}{8}} = \frac{\frac{57}{5}}{\frac{57}{8}} = \frac{57}{5}:\frac{57}{8} = \frac{8}{5} = 1\frac{3}{5} $$