Задание 2 Параграф 1 ГДЗ Рабочая тетрадь 1 Мерзляк Полонский 7 класс (Алгебра)

1) (1 1/3+2 2/9)·3/16-1 2/3·3;
2) 1 5/7 :(-6)+(1 16/21-3 4/9) :(-1 11/42);
3) (-1 1/8+27/32) :(3,05-2,84)-0,3/1,4.

1. $$\left(1\frac{1}{3}+2\frac{2}{9}\right)\cdot \frac{3}{16}-1\frac{2}{3}\cdot 3$$
   $$1\frac{1}{3}+2\frac{2}{9}=1\frac{3}{9}+2\frac{2}{9}=3\frac{5}{9}=\frac{32}{9}$$
   $$\frac{32}{9}\cdot \frac{3}{16}=\frac{2}{3}$$
   $$1\frac{2}{3}\cdot 3=\frac{5}{3}\cdot 3=5$$
   $$\frac{2}{3}-5=-4\frac{1}{3}$$
2. $$1\frac{5}{7}:(-6)+\left(1\frac{16}{21}-3\frac{4}{9}\right):\left(-1\frac{11}{42}\right)$$
   $$1\frac{5}{7}:(-6)=\frac{12}{7}:(-6)=-\frac{2}{7}$$
   $$1\frac{16}{21}-3\frac{4}{9}=\frac{37}{21}-\frac{31}{9}=\frac{111}{63}-\frac{217}{63}=-\frac{106}{63}=-1\frac{43}{63}$$
   $$-1\frac{43}{63}:\left(-1\frac{11}{42}\right)=\frac{106}{63}:\frac{53}{42}=\frac{106}{63}\cdot \frac{42}{53}=\frac{4}{3}$$
   $$-\frac{2}{7}+\frac{4}{3}=\frac{22}{21}=1\frac{1}{21}$$
3. $$\left(-1\frac{1}{8}+\frac{27}{32}\right):(3{,}05-2{,}84)-\frac{0{,}3}{1{,}4}$$
   $$-1\frac{1}{8}+\frac{27}{32}=-\frac{9}{8}+\frac{27}{32}=-\frac{36}{32}+\frac{27}{32}=-\frac{9}{32}$$
   $$3{,}05-2{,}84=0{,}21$$
   $$-\frac{9}{32}:0{,}21=-\frac{9}{32}:\frac{21}{100}=-\frac{9}{32}\cdot \frac{100}{21}=-\frac{75}{56}$$
   $$\frac{0{,}3}{1{,}4}=\frac{3}{10}:\frac{14}{10}=\frac{3}{14}$$
   $$-\frac{75}{56}-\frac{3}{14}=-\frac{75}{56}-\frac{12}{56}=-\frac{87}{56}=-1\frac{31}{56}$$

Ответ

1) $$-4\frac{1}{3}$$; 2) $$1\frac{1}{21}$$; 3) $$-1\frac{31}{56}$$.