Упр.770 ГДЗ Муравин 6 класс (Математика)

1) (152 3/4-148 3/8)•0,3:0,2;
2) (-2/3)^2•(3/4)^2•?1,5?^3•(-1 1/3)^3;
3) 6,3•(2,5•9-22,5);
4) 1,5-(3/4-5/6)+(-2 7/12).

1) $$\left(152\frac{3}{4}-148\frac{3}{8}\right)\cdot 0{,}3:0{,}2$$

$$152\frac{3}{4}=152\frac{6}{8},\quad 152\frac{6}{8}-148\frac{3}{8}=4\frac{3}{8}$$

$$4\frac{3}{8}\cdot 0{,}3:0{,}2=4\frac{3}{8}\cdot \frac{3}{10}:\frac{1}{5}=4\frac{3}{8}\cdot \frac{3}{10}\cdot 5$$

$$4\frac{3}{8}=\frac{35}{8},\quad \frac{35}{8}\cdot \frac{3}{10}\cdot 5=\frac{35\cdot 3}{8\cdot 2}=\frac{105}{16}=6\frac{9}{16}$$

2) $$\left(-\frac{2}{3}\right)^2\cdot \left(\frac{3}{4}\right)^2\cdot 1{,}5^3\cdot \left(-1\frac{1}{3}\right)^3$$

$$\left(-\frac{2}{3}\right)^2=\frac{4}{9},\quad \left(\frac{3}{4}\right)^2=\frac{9}{16},\quad 1{,}5=\frac{3}{2},\quad -1\frac{1}{3}=-\frac{4}{3}$$

$$\frac{4}{9}\cdot \frac{9}{16}\cdot \left(\frac{3}{2}\right)^3\cdot \left(-\frac{4}{3}\right)^3$$

$$=\frac{1}{4}\cdot \frac{27}{8}\cdot \left(-\frac{64}{27}\right)=-2$$

3) $$6{,}3\cdot (2{,}5\cdot 9-22{,}5)$$

$$2{,}5\cdot 9=22{,}5$$

$$6{,}3\cdot (22{,}5-22{,}5)=6{,}3\cdot 0=0$$

4) $$1{,}5-\left(\frac{3}{4}-\frac{5}{6}\right)+\left(-2\frac{7}{12}\right)$$

$$1{,}5=1\frac{1}{2},\quad \frac{3}{4}-\frac{5}{6}=\frac{9}{12}-\frac{10}{12}=-\frac{1}{12},\quad -2\frac{7}{12}=-2\frac{7}{12}$$

$$1\frac{1}{2}-\left(-\frac{1}{12}\right)-2\frac{7}{12}=1\frac{1}{2}+\frac{1}{12}-2\frac{7}{12}$$

$$1\frac{1}{2}=\frac{18}{12},\quad \frac{18}{12}+\frac{1}{12}-\frac{31}{12}=-\frac{12}{12}=-1$$

Ответ

1) $$6\frac{9}{16}$$; 2) $$-2$$; 3) $$0$$; 4) $$-1$$.