Упр.299 Часть 2 ГДЗ Дорофеев Петерсон 6 класс (Математика)

1. Вычисли: 1) $$(2{,}35\mathbin{\cdot}70{,}2)\mathbin{:}23{,}4-(38{,}36+19{,}8)\mathbin{\cdot}0{,}1$$; 2) $$16\frac{1}{3}\mathbin{:}1{,}4-\frac{0{,}75+\frac{7}{12}}{3\frac{7}{15}+0{,}1-\frac{11}{30}}\mathbin{\cdot}1{,}6$$.

1) Вычислим по порядку:

$$(2{,}35\cdot 70{,}2):23{,}4-(38{,}36+19{,}8)\cdot 0{,}1$$
$$2{,}35\cdot 70{,}2=164{,}97$$
$$164{,}97:23{,}4=7{,}05$$
$$38{,}36+19{,}8=58{,}16$$
$$58{,}16\cdot 0{,}1=5{,}816$$
$$7{,}05-5{,}816=1{,}234$$

Значит,

$$(2{,}35\cdot 70{,}2):23{,}4-(38{,}36+19{,}8)\cdot 0{,}1=1{,}234.$$

2) Преобразуем дроби и выполняем действия по порядку:

$$16\frac{1}{3}:1{,}4-\frac{0{,}75+\frac{7}{12}}{3\frac{7}{15}+0{,}1-\frac{11}{30}}\cdot 1{,}6$$
$$16\frac{1}{3}=\frac{49}{3},\qquad 1{,}4=\frac{7}{5},\qquad 0{,}75=\frac{3}{4},\qquad 0{,}1=\frac{1}{10},\qquad 1{,}6=\frac{8}{5}$$
$$\frac{49}{3}:\frac{7}{5}=\frac{49}{3}\cdot\frac{5}{7}=\frac{35}{3}$$
$$0{,}75+\frac{7}{12}=\frac{3}{4}+\frac{7}{12}=\frac{9}{12}+\frac{7}{12}=\frac{16}{12}=\frac{4}{3}$$
$$3\frac{7}{15}+0{,}1-\frac{11}{30}=3\frac{14}{30}+\frac{1}{10}-\frac{11}{30} =3+\frac{14}{30}+\frac{3}{30}-\frac{11}{30} =3+\frac{6}{30}=3+\frac{1}{5}=\frac{16}{5}$$
$$\frac{0{,}75+\frac{7}{12}}{3\frac{7}{15}+0{,}1-\frac{11}{30}} =\frac{\frac{4}{3}}{\frac{16}{5}}=\frac{4}{3}\cdot\frac{5}{16}=\frac{5}{12}$$
$$\frac{5}{12}\cdot 1{,}6=\frac{5}{12}\cdot\frac{8}{5}=\frac{2}{3}$$
$$\frac{35}{3}-\frac{2}{3}=11$$

Следовательно,

$$16\frac{1}{3}:1{,}4-\frac{0{,}75+\frac{7}{12}}{3\frac{7}{15}+0{,}1-\frac{11}{30}}\cdot 1{,}6=11.$$

Ответ

$$1)\ 1{,}234;\qquad 2)\ 11.$$