Упр.610 ГДЗ Зубарева Мордкович 6 класс ФГОС (Математика)

а) (3/4-5/6)•3+(3/4+5/6)•4;
б) (1/2+2/3) :5/6-12•(1/2-2/3);
в)-2/7•(3/5+3/10)+(3/5-3/10)•10/21;
г) (5/8-5/6) :(-15/16)-(5/8+5/6) :7/8;
д) (1 3/4-2 3/8) :1 3/5;
е) (1 7/15+3 1/3)•5/12;
ж) (4 1/8-2 5/16)•(-1 3/29);
з) (8 1/3-10 1/2)•12/65.

а) $$\left(\frac{3}{4}-\frac{5}{6}\right)\cdot 3+\left(\frac{3}{4}+\frac{5}{6}\right)\cdot 4$$

$$\frac{3}{4}-\frac{5}{6}=\frac{9-10}{12}=-\frac{1}{12}, \qquad \frac{3}{4}+\frac{5}{6}=\frac{9+10}{12}=\frac{19}{12}$$
$$-\frac{1}{12}\cdot 3+\frac{19}{12}\cdot 4=-\frac{3}{12}+\frac{76}{12}=\frac{73}{12}=6\frac{1}{12}$$

б) $$\left(\frac{1}{2}+\frac{2}{3}\right):\frac{5}{6}-12\cdot\left(\frac{1}{2}-\frac{2}{3}\right)$$

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

в) $$-\frac{2}{7}\cdot\left(\frac{3}{5}+\frac{3}{10}\right)+\left(\frac{3}{5}-\frac{3}{10}\right)\cdot\frac{10}{21}$$

$$\frac{3}{5}+\frac{3}{10}=\frac{6+3}{10}=\frac{9}{10}, \qquad \frac{3}{5}-\frac{3}{10}=\frac{6-3}{10}=\frac{3}{10}$$
$$-\frac{2}{7}\cdot\frac{9}{10}+\frac{3}{10}\cdot\frac{10}{21} =-\frac{18}{70}+\frac{3}{21} =-\frac{9}{35}+\frac{5}{35} =-\frac{4}{35}$$

г) $$\left(\frac{5}{8}-\frac{5}{6}\right):\left(-\frac{15}{16}\right)-\left(\frac{5}{8}+\frac{5}{6}\right):\frac{7}{8}$$

$$\frac{5}{8}-\frac{5}{6}=\frac{15-20}{24}=-\frac{5}{24}, \qquad \frac{5}{8}+\frac{5}{6}=\frac{15+20}{24}=\frac{35}{24}$$
$$-\frac{5}{24}:\left(-\frac{15}{16}\right)-\frac{35}{24}:\frac{7}{8} =-\frac{5}{24}\cdot\left(-\frac{16}{15}\right)-\frac{35}{24}\cdot\frac{8}{7}$$
$$=\frac{2}{9}-\frac{5}{3}=\frac{2}{9}-\frac{15}{9}=-\frac{13}{9}=-1\frac{4}{9}$$

д) $$\left(1\frac{3}{4}-2\frac{3}{8}\right):1\frac{3}{5}$$

$$1\frac{3}{4}=\frac{7}{4}, \qquad 2\frac{3}{8}=\frac{19}{8}, \qquad 1\frac{3}{5}=\frac{8}{5}$$
$$\left(\frac{7}{4}-\frac{19}{8}\right):\frac{8}{5} =\left(\frac{14}{8}-\frac{19}{8}\right):\frac{8}{5} =-\frac{5}{8}\cdot\frac{5}{8} =-\frac{25}{64}$$

е) $$\left(1\frac{7}{15}+3\frac{1}{3}\right)\cdot\frac{5}{12}$$

$$1\frac{7}{15}=\frac{22}{15}, \qquad 3\frac{1}{3}=\frac{10}{3}=\frac{50}{15}$$
$$\left(\frac{22}{15}+\frac{50}{15}\right)\cdot\frac{5}{12} =\frac{72}{15}\cdot\frac{5}{12} =\frac{24}{5}\cdot\frac{5}{12}=2$$

ж) $$\left(4\frac{1}{8}-2\frac{5}{16}\right)\cdot\left(-1\frac{3}{29}\right)$$

$$4\frac{1}{8}=\frac{33}{8}=\frac{66}{16}, \qquad 2\frac{5}{16}=\frac{37}{16}, \qquad -1\frac{3}{29}=-\frac{32}{29}$$
$$\left(\frac{66}{16}-\frac{37}{16}\right)\cdot\left(-\frac{32}{29}\right) =\frac{29}{16}\cdot\left(-\frac{32}{29}\right) =-2$$

з) $$\left(8\frac{1}{3}-10\frac{1}{2}\right)\cdot\frac{12}{65}$$

$$8\frac{1}{3}=\frac{25}{3}, \qquad 10\frac{1}{2}=\frac{21}{2}$$
$$\left(\frac{25}{3}-\frac{21}{2}\right)\cdot\frac{12}{65} =\left(\frac{50-63}{6}\right)\cdot\frac{12}{65} =-\frac{13}{6}\cdot\frac{12}{65} =-\frac{2}{5}$$

Ответ

а) $$6\frac{1}{12}$$; б) $$3\frac{2}{5}$$; в) $$-\frac{4}{35}$$; г) $$-1\frac{4}{9}$$; д) $$-\frac{25}{64}$$; е) $$2$$; ж) $$-2$$; з) $$-\frac{2}{5}$$.