Упр.308 Часть 2 ГДЗ Дорофеев Петерсон 5 класс (Математика)

Задача

1) (8 — 7 13/17) · (2 1/4 + 3 1/3 — 4 11/15) · 1;
2) (9 9/14 — 7 10/21) · 35 + (11 2/15 — 8 3/25) · 20 + (9/32 — 9/32) · 16;
3) [5 4/9 — (2 1/3)^2] · 8 1/4 · 1 3/20 + 12 · 1/3 · 3 3/4 · 1 1/5 — 4/17 · 0 · 5 18/49;
4) [(1 1/2)^3 · 4 1/6 · 8 · (1/9)^2 · 1 1/5 — 1 2/3] · 12 3/4 — 0 · 8 2/5 + 2 5/9 · 7 · 1/7.

Подробный ответ

1) $$\left(8-7\frac{13}{17}\right)\cdot\left(2\frac14+3\frac13-4\frac{11}{15}\right)\cdot 1$$

$$8-7\frac{13}{17}=7\frac{17}{17}-7\frac{13}{17}=\frac{4}{17}$$
$$2\frac14+3\frac13-4\frac{11}{15}=2\frac{15}{60}+3\frac{20}{60}-4\frac{44}{60}=5\frac{35}{60}-4\frac{44}{60}=\frac{51}{60}$$
$$\frac{4}{17}\cdot\frac{51}{60}\cdot 1=\frac{4\cdot 51}{17\cdot 60}=\frac{1}{5}$$

2) $$\left(9\frac{9}{14}-7\frac{10}{21}\right)\cdot 35+\left(11\frac{2}{15}-8\frac{3}{25}\right)\cdot 20+\left(\frac{9}{32}-\frac{9}{32}\right)\cdot 16$$

$$9\frac{9}{14}-7\frac{10}{21}=9\frac{27}{42}-7\frac{20}{42}=2\frac{7}{42}=2\frac{1}{6}$$
$$11\frac{2}{15}-8\frac{3}{25}=11\frac{10}{75}-8\frac{9}{75}=3\frac{1}{75}$$
$$\left(\frac{9}{32}-\frac{9}{32}\right)\cdot 16=0$$
$$2\frac{1}{6}\cdot 35+3\frac{1}{75}\cdot 20=70+\frac{35}{6}+60+\frac{20}{75}$$
$$=130+\frac{35}{6}+\frac{4}{15}=135+\frac{11}{10}=136\frac{1}{10}$$

3) $$\left[5\frac49-\left(2\frac13\right)^2\right]\cdot 8\frac14\cdot 1\frac{3}{20}+12\cdot\frac13\cdot 3\frac34\cdot 1\frac15-\frac{4}{17}\cdot 0\cdot 5\frac{18}{49}$$

$$\left(2\frac13\right)^2=\left(\frac73\right)^2=\frac{49}{9}$$
$$5\frac49-\frac{49}{9}=\frac{49}{9}-\frac{49}{9}=0$$
$$\frac{4}{17}\cdot 0\cdot 5\frac{18}{49}=0$$
$$12\cdot\frac13\cdot 3\frac34\cdot 1\frac15=4\cdot\frac{15}{4}\cdot\frac65=18$$
$$0\cdot 8\frac14\cdot 1\frac{3}{20}+18=18$$

4) $$\left[\left(1\frac12\right)^3\cdot 4\frac16\cdot 8\cdot\left(\frac19\right)^2\cdot 1\frac15-1\frac23\right]\cdot 12\frac34-0\cdot 8\frac25+2\frac59\cdot 7\cdot\frac17$$

$$\left(1\frac12\right)^3=\left(\frac32\right)^3=\frac{27}{8},\quad 4\frac16=\frac{25}{6},\quad \left(\frac19\right)^2=\frac{1}{81},\quad 1\frac15=\frac65$$
$$\frac{27}{8}\cdot\frac{25}{6}\cdot 8\cdot\frac{1}{81}\cdot\frac65=1$$
$$1-1\frac23=1-\frac53=-\frac23$$
$$-\frac23+1\frac23=0$$
$$0\cdot 12\frac34=0,\qquad 0\cdot 8\frac25=0,\qquad 2\frac59\cdot 7\cdot\frac17=2\frac59$$
$$0+2\frac59=2\frac59$$