Упр.106 Часть 2 ГДЗ Дорофеев Петерсон 6 класс (Математика)

Задача

1) 2 1/7 :(1,5a)=5/14 :0,8;
2) 4,8/(1 7/9)=(1,2b) :6 2/3;
3) 3,8:(4c+3)=2:2 2/19;
4) (0,3d-1,5)/(3 1/3)=0,84:7/15;
5) (3x-1)/5=(x+1)/3;
6) 0,5y/(4/9)=(y+3)/8.

Подробный ответ

1) $$2\frac{1}{7}:(1{,}5a)=\frac{5}{14}:0{,}8$$

$$1{,}5a\cdot \frac{5}{14}=2\frac{1}{7}\cdot 0{,}8$$
$$\frac{3}{2}a\cdot \frac{5}{14}=\frac{15}{7}\cdot \frac{4}{5}$$
$$a=\frac{15\cdot 4\cdot 2\cdot 14}{7\cdot 5\cdot 3\cdot 5}=\frac{16}{5}=3{,}2$$

2) $$\frac{4{,}8}{1\frac{7}{9}}=(1{,}2b):6\frac{2}{3}$$

$$1\frac{7}{9}\cdot 1{,}2b=4{,}8\cdot 6\frac{2}{3}$$
$$\frac{16}{9}\cdot \frac{12}{10}b=\frac{48}{10}\cdot \frac{20}{3}$$
$$b=\frac{48\cdot 20\cdot 9\cdot 10}{10\cdot 3\cdot 16\cdot 12}=15$$

3) $$3{,}8:(4c+3)=2:2\frac{2}{19}$$

$$2(4c+3)=3{,}8\cdot 2\frac{2}{19}$$
$$8c+6=\frac{38}{10}\cdot \frac{40}{19}$$
$$8c+6=8$$
$$8c=2$$
$$c=\frac{2}{8}=\frac{1}{4}=0{,}25$$

4) $$\frac{0{,}3d-1{,}5}{3\frac{1}{3}}=0{,}84:\frac{7}{15}$$

$$0{,}3d-1{,}5=3\frac{1}{3}\cdot 0{,}84:\frac{7}{15}$$
$$0{,}3d-1{,}5=\frac{10}{3}\cdot \frac{84}{100}:\frac{7}{15}$$
$$0{,}3d-1{,}5=6$$
$$0{,}3d=7{,}5$$
$$d=25$$

5) $$\frac{3x-1}{5}=\frac{x+1}{3}$$

$$3(3x-1)=5(x+1)$$
$$9x-3=5x+5$$
$$9x-5x=5+3$$
$$4x=8$$
$$x=2$$

6) $$\frac{0{,}5y}{\frac{4}{9}}=\frac{y+3}{8}$$

$$0{,}5y\cdot 8=\frac{4}{9}(y+3)$$
$$4y=\frac{4}{9}y+\frac{4}{3}$$
$$4y-\frac{4}{9}y=\frac{4}{3}$$
$$3\frac{5}{9}y=\frac{4}{3}$$
$$y=\frac{4}{3}:\frac{32}{9}=\frac{4}{3}\cdot \frac{9}{32}=\frac{3}{8}$$