Упр.1.7 Часть 1 ГДЗ Мордкович Семенов 7 класс (Алгебра)

1. а)

   $$2\frac{2}{9}+2\frac{2}{7}\cdot 0{,}3+2\frac{2}{7}\cdot \frac{2}{5}+5\frac{7}{9}-8\frac{3}{5}$$
   $$=\left(2\frac{2}{9}+5\frac{7}{9}\right)+2\frac{2}{7}\left(0{,}3+\frac{2}{5}\right)-8\frac{3}{5}$$
   $$=8+\frac{16}{7}\cdot 0{,}7-8\frac{3}{5}$$
   $$=8+1{,}6-8{,}6=1.$$

2. б)

   $$4\left(1\frac{3}{4}-2\frac{1}{12}\right)-3\frac{7}{16}+\left(2\frac{1}{3}+1\frac{1}{16}\right)$$
   $$=4\cdot \frac{7}{4}-4\cdot \frac{25}{12}-\left(3\frac{7}{16}-1\frac{1}{16}\right)+2\frac{1}{3}$$
   $$=7-\frac{25}{3}-2\frac{6}{16}+2\frac{1}{3}$$
   $$=\left(7-2\frac{3}{8}\right)-\left(8\frac{1}{3}-2\frac{1}{3}\right)$$
   $$=4\frac{5}{8}-6=-1\frac{3}{8}.$$

3. в)

   $$3\frac{5}{12}+3\frac{1}{9}\cdot 1{,}5-3\frac{1}{9}\cdot \frac{6}{7}+2\frac{1}{12}-3\frac{5}{7}$$
   $$=\left(3\frac{5}{12}+2\frac{1}{12}\right)+3\frac{1}{9}\left(1{,}5-\frac{6}{7}\right)-3\frac{5}{7}$$
   $$=5\frac{1}{2}+\frac{28}{9}\cdot \left(\frac{3}{2}-\frac{6}{7}\right)-3\frac{5}{7}$$
   $$=5\frac{1}{2}+\frac{28}{9}\cdot \frac{9}{14}-3\frac{5}{7}$$
   $$=5\frac{1}{2}+2-3\frac{5}{7}=7\frac{1}{2}-3\frac{5}{7}=3\frac{11}{14}.$$

4. г)

   $$5\frac{6}{11}-\left(2\frac{7}{15}-1\frac{3}{4}\right)\cdot 3+\left(5\frac{5}{11}-4{,}5\right)+3\frac{9}{20}$$
   $$=5\frac{6}{11}-\left(\frac{37}{15}-\frac{7}{4}\right)\cdot 3+5\frac{5}{11}-4{,}5+3\frac{9}{20}$$
   $$=11-\left(7{,}4-4{,}5\right)+3{,}45$$
   $$=11-2{,}9+3{,}45=11+0{,}55=11{,}55.$$

Ответ

а) $$1$$; б) $$-1\frac{3}{8}$$; в) $$3\frac{11}{14}$$; г) $$11{,}55$$.